arXiv:1702.00284v1 [math.CO] 31 Jan 2017
Musical intervals under 12-note equal
temperament: a geometrical interpretation
R. Caimmi∗†, M◦ A. Franzon‡§, S. Tognon¶
February 2, 2017
Abstract
Musical intervals in multiple of semitones under 12-note equal temperament, or more specifically pitch-class subsets of assigned cardinality,
n, 1 ≤ n ≤ 12, (n-chords) are conceived as positive integer points,
Pn ≡ (ℓ1 , ℓ2 , ..., ℓn ), ℓ1 + ℓ2 + ... + ℓn = 12, within an Euclidean nspace, ℜn . The number of distinct n-chords, NC (n), is inferred from
combinatorics with the extension to n = 0, involving an Euclidean
0-space, ℜ0 . The number of repeating n-chords, ∆N (n), or points
which are turned into themselves during a circular permutation, Tn ,
of their coordinates, is inferred from algebraic considerations. Finally, the total number of n-chords, NM (n) = NC (n) + ∆N (n), and
the number of Tn set classes, νM (n) = NM (n)/n, are determined.
Palindrome and pseudo palindrome n-chords are defined and included
among repeating n-chords, with regard to an equivalence relation,
Tn /Tn I, where reflection is added to circular permutation. To this respect, the number of Tn set classes is inferred concerning palindrome
∗
Physics and Astronomy Department, Padua University, Vicolo Osservatorio 3/2,
35122 Padova, Italy email: [email protected] fax: 39-049-8278212
†
Affiliated up to September 30th 2014. Current status: Studioso Senior. Current
position: in retirement due to age limits.
‡
Associazione Culturale S. Nicolò, Piazza Pio X 27, 36043 Camisano Vicentino (VI),
Italy email: [email protected]
§
Via Fogazzaro 32, 36043 Camisano Vicentino (VI), Italy
¶
FBD House, Bluebell, Dublin 12, Ireland email: [email protected]
fax:+353
(0) 1 478 3574
1
and pseudo palindrome n-chords, νP (n), and the remaining n-chords,
νN (n) = νM (n) − νP (n), yielding a number of Tn /Tn I set classes,
νQ (n) = νN (n)/2 + νP (n). The above results are reproduced within
the framework of a geometrical interpretation, where positive integer points related to n-chords of cardinality, n, belong to a regular
inclined n-hedron, Ψn12 , the vertexes lying on the coordinate axes of
a Cartesian orthogonal reference frame, (Ox1 x2 ...xn ), at a distance,
xi = 12, 1 ≤ i ≤ n, from the origin. Considering Ψn12 as special cases
of lattice polytopes, the number of related nonnegative integer points
is also determined for completeness. A comparison is performed with
the results inferred from group theory. The symmetry of the number
of n-chords, Tn set classes, Tn /Tn I set classes, with regard to cardinality, is interpreted as intrinsic to n-hedrons and, for this reason,
expressed via the binomial formula. More generally, the symmetry
of the results inferred from the group theory could be conceived as
intrinsic to lattice polytopes in ℜn .
Keywords: pitch-classes; n-chords; Tn set classes; Tn /Tn I set classes;
Euclidean n-spaces; n-hedrons
1
Introduction
The question of how many musical intervals in multiples of semitones there
are under 12-note equal temperament (more specifically, how many pitchclass subsets there are of a given cardinality, or how many there are with
respect to one fixed pitch-class and further organized under various equivalence relations) is one that has been answered early and often in the music
theory and mathematical literature e.g., [1][2]. The problem has been worked
out independently, in particular the partition into equivalence classes under
transposition (or circular permutation) e.g., [3][4][5]. To this respect, it is
worth emphasyzing combinatorial problems are not essentially musical: the
same procedure can be applied e.g., for the isomer enumeration in chemistry,
for spin analysis in physics, and in general for the investigation of isomorphism classes of objects e.g., [6][7]. The most elegant way for solving such
problems is the Polya-Burnside method, which was applied to music theory
more than thirty years ago [8].
Both pitch-class subsets and Tn set classes of each cardinality from 0
through 12 are familiar to music theorists but the set-class counts, in absence
of a deep knowledge of the group theory, are performed by use of tables enumerating all the set classes e.g., [9][3][4][10][11]. An intermediate use between
the two extremes mentioned above relates to standard techniques in classical
combinatorial theory and offers some simple applications to music theory, in2
cluding the enumeration of pitch-class subsets e.g., [12][13][15]. In addition,
following this line of thought foreshadows certain aspects of the more difficult work involved in group theory, and therefore may form a pedagogically
benefical bridge to the advanced material e.g., [14] Chap. 9 [15][17].
To this respect, the present paper is restricted to the simplest case of Tn
and Tn /Tn I set classes, with regard to pitch-class subsets (internal patterns
or internal structures) of cardinality, n, where the sum of musical intervals in
multiples of semitones equals 12, or n-chords. Of course, related results are
already known in the literature e.g., [8][14] Chap. 9 [15][16][17][18][19], but
the exposition here is expected to be more readily accessible to music theory
community and, last but not least, to interest in group theory by itself. The
current approach is essentially algebraic and geometric: in short, the paper
presents “an algorithmic theory,” one of many possible. The main steps of
the method may be summarized as follows.
First, n-chords, {ℓ1 , ℓ2 , ...ℓn }, are conceived as positive integer points of
coordinates, Pn ≡ (ℓ1 , ℓ2 , ...ℓn ), with respect to a Cartesian orthogonal reference frame, (Ox1 x2 ...xn ), in an Euclidean n-space, ℜn . In this view, n-chords
may be thought of as made of coordinates. Then Tn set classes of each cardinality are partitioned into two main categories, namely set classes where
n-chords exhibit distinct e.g., (1,2,3,6) and repeating e.g., (2,4,2,4) coordinates, respectively.
Second, n-chords belonging to the above mentioned categories are enumerated separately and the amount of related Tn set classes is determined.
Third, the number of Tn /Tn I set classes of each cardinality is also determined following a similar procedure.
Fourth, further attention is devoted to the geometrical interpretation in
itself.
The method could, in principle, be extended to musical intervals in multiples of semitones under L-note (instead of 12-note) equal temperament.
The text is organized as follows. The first, second and third step outlined
above are developed in different subsections of Section 2. The fourth step
is considered in Section 3. The discussion is presented in Section 4. The
conclusion is shown in Section 5. As guidance examples, the method is
applied to classical birthday-cake and necklace problem in Appendix A and
B, respectively. General properties of n-hedrons are outlined in Appendix C.
3
2
Enumeration of n-chords, Tn and Tn /Tn I set
classes
A pitch-class subset is defined to be a subset of the set of twelve pitch-classes
e.g., [14] Chap. 9 §9.14. In musical terms, natural numbers within the range,
1 ≤ n ≤ 12, could be thought of as representing musical intervals in multiples
of semitones, in the twelve tone equal tempered octave. Octave equivalence in
the musical scale implies two notes belong to the same pitch-class if they differ
by a whole number of octaves. Then addition has an obvious interpretation
as addition of musical intervals. To this respect, an origin must be chosen via
one fixed pitch-class. For further details, an interested reader is addressed to
specific investigations e.g., [17] or textbooks e.g., [14] Chap. 9.
Accordingly, pitch-class subsets of cardinality, n, are denoted as n-tuples,
{ℓ1 , ℓ2 , ..., ℓn }, where ℓ1 , ℓ2 , ..., ℓn , are natural numbers. Let n-chords be
defined as pitch-class subsets where the boundary condition:
ℓ1 + ℓ2 + ... + ℓn = 12 ;
1 ≤ n ≤ 12 ;
(1)
is satisfied e.g., [17].
Tn set classes are obtained by transposition (or circular permutation) as
{ℓ1 , ℓ2 , ..., ℓn }, {ℓ2 , ℓ3 , ..., ℓ1 }, ..., {ℓn , ℓ1 , ..., ℓn−1 }. Tn /Tn I set classes are obtained by reflection (or order inversion) followed by a transposition. More
specifically, the application of the pitch-class operator, Tk , k ≤ n, on the ntuple, {ℓ1 , ℓ2 , ..., ℓn }, yields {ℓk+1, ..., ℓn , ℓ1 , ..., ℓk }, and the application of the
pitch-class operator, Tk I, on the same n-tuple, yields a reflection, {ℓn , ℓn−1 , ..., ℓ1 },
followed by a transposition, {ℓk , ..., ℓ1 , ℓn , ..., ℓk+1 }.
Let the prime form of a set class be defined as a special n-chord within
that class, for which (i) the last element of the n-tuple has the larger value
and, in case of multiplicity, (ii) the first element of the n-tuple has the lower
value [20]. For instance, the prime form of the Tn /Tn I set class,
{1, 2, 9}, {2, 9, 1}, {9, 1, 2}, {1, 9, 2}, {9, 2, 1}, {2, 1, 9};
is {1, 2, 9}. Accordingly, Tn and Tn /Tn I set classes can be represented by
their prime forms, as cyclic “adjacent internal array” (CINT1 ) [20].
Let n-chords be defined as “distinct” and “repeating” according if related
n-tuples are different or coinciding, respectively. Let Tn set classes be defined
as “distinct” and “repeating” according if related n-chords are distinct or
repeating, respectively. For instance, the Tn set class,
{1, 2, 3, 6}, {2, 3, 6, 1}, {3, 6, 1, 2}, {6, 1, 2, 3};
4
is made of four distinct 4-chords, while the Tn set class,
{1, 5, 1, 5}, {5, 1, 5, 1}, {1, 5, 1, 5}, {5, 1, 5, 1};
is made of two distinct and two repeating 4-chords. It is worth mentioning
repeating musical intervals in multiples of semitones have been studied and
used since about 70 years ago [12].
A method shall be exploited in the following subsections, where distinct
and repeating n-chords shall be counted separately to yield the number of
Tn and Tn /Tn I set classes.
2.1
Enumeration of distinct n-chords
Aiming to a geometrical interpretation, n-tuples representing n-chords shall
be considered as coordinates of points, Pn ≡ (ℓ1 , ℓ2 , ..., ℓn ), with respect to
a Cartesian orthogonal reference frame, (O x1 x2 ... xn ), in an Euclidean ndimension hyperspace, or n-space, ℜn . More specifically, Pn lies within the
positive 2n -ant (2-ant is versant, 4-ant is quadrant, 8-ant is octant, and so
on), and the coordinates are natural numbers linked via Eq. (1).
With no loss of generality, the dependent coordinate may be chosen to
be ℓn , as:
ℓn = 12 − ℓ1 − ℓ2 − ... − ℓn−1 ;
1 ≤ n ≤ 12 ;
(2)
and the projection of Pn onto the principal (n − 1)-dimension hyperplane, or
(n − 1)-plane, (Ox1 x2 ... xn−1 ), is Pn−1 ≡ (ℓ1 , ℓ2 , ..., ℓn−1 ). The knowledge of
Pn−1 implies the knowledge of Pn via Eq. (2).
Given a generic projected point, Pn−1 ≡ (ℓ1 , ℓ2 , ..., ℓn−2 , ℓn−1 ), let the conjugate point be defined as Qn−1 ≡ (12 − ℓ1 , S2,n−1, ..., Sn−2,n−1 , ℓn−1), where,
in general, Si,n−k are expressed as:
Si,n−k = ℓi + ℓi+1 + ... + ℓn−k−1 + ℓn−k ;
0 < ℓn−1 < Sn−2,n−1 < ... < S3,n−1 < S2,n−1 < 12 − ℓ1 ;
0 < ℓi < 12 ;
1≤i≤n−1 ;
(3a)
(3b)
(3c)
and Eq. (3b) follows from (1), (3c).
According to Eq. (3), the projected points, Pn−1 , Qn−1 , are in a 1 : 1
correspondence, Pn−1 ↔ Qn−1 , or:
(ℓ1 , ℓ2 , ..., ℓn−2, ℓn−1 ) ↔ (12 − ℓ1 , S2,n−1, ..., Sn−2,n−1 , ℓn−1 ) ;
(4)
where the coordinates on the right-hand side of Eq. (4) are clearly distinct,
monotonically decreasing, and belonging to the subset of natural numbers,
{1, 2, ..., 11}, via Eq. (3). The special case, n = 3, is shown in Fig. 1.
5
With regard to the projected point, Qn−1 , there are 11 − n + n = 11 − 0
different ways of choosing the first coordinate, 11 − n + (n − 1) = 11 − 1
different ways of choosing the second coordinate with the preceeding fixed,
..., 11 − n + [n − (n − 2)] = 11 − n + 2 different ways of choosing the (n − 1)th
coordinate with the preceeding fixed, for a total, NC′ = 11 · 10 · ... · (11 − n +
2) = 11!/[11 − (n − 1)]!, including points whose coordinates are linked by
permutations i.e. with place exchanged one with respect to the other.
For (n − 1) fixed distinct coordinates, there are (n − 1) different ways of
choosing the first coordinate, (n − 2) different ways of choosing the second
coordinate with the preceeding fixed, ..., [n − (n − 2)] = 2 different ways of
choosing the (n − 2)th coordinate with the preceeding fixed, [n − (n − 1)] = 1
univocal way of choosing the (n − 1)th coordinate with the preceeding fixed,
for a total, N ′ = (n − 1) · (n − 2) · ... · 2 · 1 = (n − 1)!.
In conclusion, the total number of projected points, Qn−1 , having coordinates (i) belonging to the subset of natural numbers, {1, 2, ..., 11}; (ii) distinct the one with respect to the other; (iii) univocally ordered i.e. excluding
permutations between coordinates; is expressed by the ratio, NC = NC′ /N ′ ,
as:
!
11!
1
n
12!
n 12
NC =
;
(5)
=
=
[11 − (n − 1)]! (n − 1)!
12 (12 − n)!n!
12 n
in terms of the binomial coefficients:
N
K
!
N!
N!
N
=
=
=
K!(N − K)!
(N − K)!K!
N −K
!
;
(6)
related to any pair of natural numbers, N, K, N ≥ K.
Accordingly, NC is the number of distinct n-chords of cardinality, n. On
the other hand, the total number of pitch-class sets of cardinality, n, regardless of Eq. (1), reads 12NC /n e.g., [15] §29. Owing to Eq. (6), the dependence
of NC on n is symmetric, as shown in Table 1. The additional case, n = 0,
has been added for completing the symmetry and shall be considered below. Further symmetries are exhibited by the fractional number of distinct
n-chords, NC /n, and the related integer part, IC = Int(NC /n), via Eq. (6).
More specifically, Eq. (5) via (6) takes the equivalent form:
!
!
NC
1
12
1 12
=
=
n
12 n
12 12 − n
;
(7)
and the related integer part is:
NC
IC = Int
n
!#
"
1 12
= Int
12 n
6
"
!#
1
12
= Int
12 12 − n
;
(8)
Table 1: Number and fractional number of distinct n-chords, NC , NC /n,
integer part, IC = Int(NC /n), number and fractional number of repeating
n-chords, ∆N, ∆N/n, total number and fractional total number of n-chords,
NM = NC + ∆N, NM /n, for different cardinality, n, 1 ≤ n ≤ 12. The
additional case, n = 0, has been added for completing the symmetry. See
text for further details.
n
NC
NC
n
IC
∆N
∆N
n
NM
NM
n
0
0
1
12
0
0
11
12
0
1
1
1
1
1
0
0
1
1
2
11
11
2
5
1
1
2
12
6
3
55
55
3
18
2
2
3
57
19
4
165
165
4
41
7
7
4
172
43
5
6
7
330 462 462
66 77 66
66 77 66
0
18
0
0
3
0
330 480 462
66 80 66
8
330
165
4
41
14
7
4
344
43
9
165
55
3
18
6
2
3
171
19
10
55
11
2
5
5
1
2
60
6
11
11
1
1
0
0
11
1
12
1
1
12
0
11
11
12
12
1
which is symmetric with respect to the maximum, occurring at n = 6, as
shown in Table 1.
To complete the symmetry up to the extreme value, n = 12, the domain must be extended down to the opposite extreme, n = 0, conceived as
representing the empty n-chord (no mode). To this aim, factorials must be
expressed in terms of the Euler Gamma function e.g., [21] Chap. 16, as:
Γ(n + 1) = nΓ(n) = n! ;
Γ(1) = 1 ;
n = 1, 2, 3, ... ;
(9)
where the recursion formula holds for all positive reals, in particular:
lim [nΓ(n)] = lim+ Γ(n + 1) = Γ(1) = 1 ;
n→0+
(10)
n→0
or:
1
;
(11)
n→0
n→0 n
and a similar result is found for n → 0− extending the recursion formula,
Eq. (9), to the negative real semiaxis.
In terms of the Euler Gamma function, Eq. (7) reads:
lim+ Γ(n) = lim+
1
Γ(12)
NC
=
;
n
n Γ(13 − n)Γ(n)
(12)
which, for positive infinitesimal n, takes the expression:
#
"
"
NC
Γ(12)
Γ(12)
1
1 1
lim+
=
= lim+
lim+
n→0
n→0
n
n Γ(13 − n)Γ(n)
Γ(13) n→0 n Γ(n)
7
#
;
(13)
and the combination of Eqs. (9), (10), (13), yields:
NC
11!
1
=
=
;
n→0
n
12!
12
NC
lim Int
=0 ;
n→0+
n
lim+
(14)
(15)
which completes the symmetry of the fractional number, NC /n, and related
integer part, IC , with respect to the maximum occurring at n = 6. In authors’
opinion, the above considerations add something more to the bare statement,
that 0! = 1 holds by definition.
With regard to a selected Euclidean n-space, an integer value of the
fractional number, NC /n, makes a necessary (but not sufficient) condition
for a one-to-one correspondence between projected points and coordinates,
Qn−1 ↔ {s1 , s2 , ..., sn−1 }, where s1 = 12 − ℓ1 ; sk = Sk,n−1 , 2 ≤ k ≤ n − 2;
sn−1 = ℓn−1 . An inspection of Table 1 shows the necessary condition fails
in several cases, which implies the above mentioned correspondence is not
one-to-one i.e. repeating coordinates, related to repeating n-chords, must
also be enumerated.
2.2
Enumeration of repeating n-chords
With regard to a selected Euclidean n-space and a primitive form, Pn ≡
(ℓ1 , ℓ2 , ..., ℓn ), of an assigned Tn set class, repeating (or transpositionally invariant) n-chords, P′n ≡ (ℓ′1 , ℓ′2 , ..., ℓ′n ), P′′n ≡ (ℓ′′1 , ℓ′′2 , ..., ℓ′′n ), exhibit identical
coordinates, ℓ′j = ℓ′′j , 1 ≤ j ≤ n. More specifically, a necessary condition for
the occurrence of repeating n-chords is that the coordinates, (ℓ1 , ℓ2 , ..., ℓn ),
equal one to the other in the same number i.e. ℓ11 = ℓ21 = ... = ℓi1 ;
ℓ12 = ℓ22 = ... = ℓi2 ; ...; ℓ1k = ℓ2k = ... = ℓik ; where 1 ≤ ik = n ≤ 12.
Accordingly, repeating n-chords exhibit identical soloes of coordinates,
(ℓ1 , ℓ1 , ..., ℓ1 ), or identical duoes of coordinates, (ℓ1 , ℓ2 , ℓ1 , ℓ2 , ..., ℓ1 , ℓ2 ), or
identical trioes of coordinates, (ℓ1 , ℓ2 , ℓ3 , ℓ1 , ℓ2 , ℓ3 , ..., ℓ1 , ℓ2 , ℓ3 ), or identical
quartets of coordinates, (ℓ1 , ℓ2 , ℓ3 , ℓ4 , ℓ1 , ℓ2 , ℓ3 , ℓ4 , ℓ1 , ℓ2 , ℓ3 , ℓ4 ), or identical quintets of coordinates, (ℓ1 , ℓ2 , ℓ3 , ℓ4 , ℓ5 , ℓ1 , ℓ2 , ℓ3 , ℓ4 , ℓ5 ). Identical sextets are not
considered in that they yield the chromatic scale, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
via Eq. (1), and thus reduce to identical soloes.
For repeating n-chords of cardinality, n, Eq. (1) reduces to:
ℓ1 + ℓ2 + ... + ℓi =
12
12
= i ;
k
n
(16)
where i is the number of different coordinates and k = n/i is their multiplicity. In any case, the number of repeating n-chords, ∆Ni (n), to be added
8
to the number of distinct n-chords, NC (n), has to be determined for Tn set
classes, while Tn /Tn I set classes shall be considered afterwards.
Identical soloes of coordinates (i = 1, k = n) cannot occur for n < 2, and
Eq. (16) reduces to:
12
ℓ1 =
;
n≥2 ;
(17)
n
which implies the existence of n-chords with identical soloes of coordinates
provided the ratio on the right-hand side of Eq. (17) is integer. The related
Tn set class is made of n identical singletons of n-chords, one to be counted
as distinct and the remaining (n − 1) to be added as repeating. Accordingly,
the number of repeating n-chords reads:
∆N1 (n) = ζ(12, n)(n − 1)ν1 (n) ;
n≥2 ;
(18)
where ν1 (n) is the number of Tn set classes including n-chords which satisfy
Eq. (17), more specifically ν1 (n) = 1 for n = 2, 3, 4, 6, 12, and ν1 (n) = 0
otherwise. In general the function, ζ, is defined as:



1
;
ζ(m1 , m2 ) = 1 ;


0 ;
m1 − Int m1 = 0 ;
m2
m2
m2 = 0 ; m1
m1
m − Int m > 0 ;
2
(19)
2
where Int(x) is the integer part of x and m1 , m2 , m1 ≥ m2 , are natural
numbers, m1 = 12, m2 = n, in the case under discussion.
Identical duoes of coordinates (i = 2, k = n/2) cannot occur for n < 4,
and Eq. (16) reduces to:
ℓ1 + ℓ2 =
24
;
n
n≥4 ;
(20)
which implies the existence of n-chords with identical duoes of coordinates
provided the ratio on the right-hand side of Eq. (20) is integer. Related Tn
set classes are made of n/2 identical doublets of n-chords, each one to be
counted as distinct and the others to be added as repeating. Accordingly,
the number of repeating n-chords reads:
∆N2 (n) = ζ(24, n)(n − 2)ν2 (n) ;
n≥4 ;
(21)
where ν2 (n) is the number of Tn set classes including n-chords which satisfy
Eq. (20), to be determined for n = 4, 6, 8, as ν2 (n) = 0 otherwise.
For n = 4, Eq. (20) reduces to ℓ1 + ℓ2 = 6 which has distinct (i.e. at least
one different from the other) solutions as (ℓ1 , ℓ2 ) = (1, 5), (2, 4), implying
ν2 (4) = 1 + 1 = 2, ∆N2 (4) = 2 + 2 = 4.
9
For n = 6, Eq. (20) reduces to ℓ1 + ℓ2 = 4 which has distinct solutions as
(ℓ1 , ℓ2 ) = (1, 3), implying ν2 (6) = 1, ∆N2 (6) = 2 + 2 = 4.
For n = 8, Eq. (20) reduces to ℓ1 + ℓ2 = 3 which has distinct solutions
as (ℓ1 , ℓ2 ) = (1, 2), implying ν2 (8) = 1, ∆N2 (8) = 3 + 3 = 6. The 8-chord,
{1, 2, 1, 2, 1, 2, 1, 2}, is quoted among modes à transpositions limitées [12].
Identical trioes of coordinates (i = 3, k = n/3) cannot occur for n < 6,
and Eq. (16) reduces to:
ℓ1 + ℓ2 + ℓ3 =
36
;
n
n≥6 ;
(22)
which implies the existence of n-chords with identical trioes of coordinates
provided the ratio on the right-hand side of Eq. (22) is integer. Related Tn set
classes are made of n/3 identical triplets of n-chords, each one to be counted
as distinct and the others to be added as repeating. Accordingly, the number
of repeating n-chords reads:
∆N3 (n) = ζ(36, n)(n − 3)ν3 (n) ;
n≥6 ;
(23)
where ν3 (n) is the number of Tn set classes including n-chords which satisfy
Eq. (22), to be determined for n = 6, 9, as ν3 (n) = 0 otherwise.
For n = 6, Eq. (22) reduces to ℓ1 + ℓ2 + ℓ3 = 6 which has distinct solutions
as (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 3), (1, 1, 4), implying ν3 (6) = 2+1 = 3, ∆N3 (6) = 3+3+
3 = 9. The 6-chord, {1, 4, 1, 1, 4, 1}, is quoted among modes à transpositions
limitées [12].
For n = 9, Eq. (22) reduces to ℓ1 + ℓ2 + ℓ3 = 4 which has distinct solutions
as (ℓ1 , ℓ2 , ℓ3 ) = (1, 1, 2), implying ν3 (9) = 1, ∆N3 (9) = 2 + 2 + 2 = 6. The 9chord, {2, 1, 1, 2, 1, 1, 2, 1, 1}, is quoted among modes à transpositions limitées
[12].
Identical quartets of coordinates (i = 4, k = n/4) cannot occur for n < 8,
and Eq. (16) reduces to:
ℓ1 + ℓ2 + ℓ3 + ℓ4 =
48
;
n
n≥8 ;
(24)
which implies the existence of n-chords with identical quartets of coordinates
provided the ratio on the right-hand side of Eq. (24) is integer. Related Tn
set classes are made of n/4 identical quadruplets of n-chords, each one to be
counted as distinct and the others to be added as repeating. Accordingly,
the number of repeating n-chords reads:
∆N4 (n) = ζ(48, n)(n − 4)ν4 (n) ;
n≥8 ;
(25)
where ν4 (n) is the number of Tn set classes including n-chords which satisfy
Eq. (24), to be determined for n = 8, as ν4 (n) = 0 otherwise.
10
For n = 8, Eq. (24) reduces to ℓ1 + ℓ2 + ℓ3 + ℓ4 = 6 which has distinct
solutions as (ℓ1 , ℓ2 , ℓ3 , ℓ4 ) = (1, 1, 2, 2), (1, 1, 1, 3), implying ν4 (8) = 1 + 1 = 2,
∆N4 (8) = 4 + 4 = 8. The 8-chords, {2, 2, 1, 1, 2, 2, 1, 1}, {1, 1, 3, 1, 1, 1, 3, 1},
are quoted among modes à transpositions limitées [12].
Identical quintets of coordinates (i = 5, k = n/5) cannot occur for n < 10,
and Eq. (16) reduces to:
ℓ1 + ℓ2 + ℓ3 + ℓ4 + ℓ5 =
60
;
n
n ≥ 10 ;
(26)
which implies the existence of n-chords with identical quintets of coordinates
provided the ratio on the right-hand side of Eq. (26) is integer. Related Tn
set classes are made of n/5 identical quintuplets of n-chords, each one to be
counted as distinct and the others to be added as repeating. Accordingly,
the number of repeating n-chords reads:
∆N5 (n) = ζ(60, n)(n − 5)ν5 (n) ;
n ≥ 10 ;
(27)
where ν5 (n) is the number of Tn set classes including n-chords which satisfy
Eq. (26), to be determined for n = 10, as ν5 (n) = 0 otherwise.
For n = 10, Eq. (26) reduces to ℓ1 + ℓ2 + ℓ3 + ℓ4 + ℓ5 = 6 which has
distinct solutions as (ℓ1 , ℓ2 , ℓ3 , ℓ4 , ℓ5 ) = (1, 1, 1, 1, 2), implying ν5 (10) = 1,
∆N5 (10) = 5. The 10-chord, {1, 1, 1, 2, 1, 1, 1, 1, 2, 1}, is quoted among modes
à transpositions limitées [12].
In summary, the number of repeating n-chords of each cardinality, with
regard to Tn set classes, reads:
∆N(n) =
5
X
i=1
∆Ni (n) =
5
X
i=1
ζ(12i, n)(n − i)νi (n) ;
(28)
where the number of repeating n-chords including identical soloes (i = 1),
∆N1 , duoes (i = 2), ∆N2 , trioes (i = 3), ∆N3 , quartets (i = 4), ∆N4 ,
P
quintets (i = 5), ∆N5 , and the total, ∆N = i ∆Ni , are listed in Table 2.
Similarly, the number of Tn set classes including repeating n-chords, νi ,
P
the total number of Tn set classes including repeating n-chords, ν = i νi ,
the number of Tn set classes including only distinct n-chords, νC , and the
total number of Tn set classes, νM = NM /n, are listed in Table 3.
The number of distinct + repeating n-chords of each cardinality, with
respect to Tn set classes, via Eqs. (5) and (28) reads:
!
5
X
n 12
NM (n) = NC (n) + ∆N(n) =
+
ζ(12i, n)(n − i)νi (n) ;
12 n
i=1
11
(29)
Table 2: Number of repeating n-chords of each cardinality, n, including
identical singletons, ∆N1 ; doublets, ∆N2 ; triplets, ∆N3 ; quadruplets, ∆N4 ;
P
quintuplets, ∆N5 ; total number of repeating n-chords, ∆N = i ∆Ni ; total
number of distinct n-chords, NC ; and total number of distinct + repeating
n-chords, NM = NC + ∆N. See text for further details.
n
∆N1
∆N2
∆N3
∆N4
∆N5
∆N
NC
NM
0
0
0
0
0
0
0
0
0
1 2 3
4
5
6
7
8
9 10 11 12
0 1 2
3
0
5
0
0
0 0 0 11
0 0 0
4
0
4
0
6
0 0 0 0
0 0 0
0
0
9
0
0
6 0 0 0
0 0 0
0
0
0
0
8
0 0 0 0
0 0 0
0
0
0
0
0
0 5 0 0
0 1 2
7
0 18
0 14
6 5 0 11
1 11 55 165 330 462 462 330 165 55 11 1
1 12 57 172 330 480 462 344 171 60 11 12
Table 3: Number of Tn set classes including repeating n-chords of each cardinality, n, made of identical soloes, ν1 ; duoes, ν2 ; trioes, ν3 ; quartets, ν4 ;
quintets, ν5 ; total number of Tn set classes including repeating n-chords,
P
ν = i νi ; total number of Tn set classes including only distinct n-chords,
νC ; and total number of Tn set classes including distinct + repeating n-chords,
νM = νC + ν. The T0 set class has been arbitrarily conceived as including
0-chords made of (no) identical soloes, to preserve symmetry in νC and ν.
See text for further details.
n
ν1
ν2
ν3
ν4
ν5
ν
νC
νM
0
1
0
0
0
0
1
0
1
1
0
0
0
0
0
0
1
1
2 3 4 5
1 1 1 0
0 0 2 0
0 0 0 0
0 0 0 0
0 0 0 0
1 1 3 0
5 18 40 66
6 19 43 66
6 7 8 9 10
1 0 0 0 0
1 0 1 0 0
3 0 0 1 0
0 0 2 0 0
0 0 0 0 1
5 0 3 1 1
75 66 40 18 5
80 66 43 19 6
12
11 12
0 1
0 0
0 0
0 0
0 0
0 1
1 0
1 1
and the number of Tn set classes of each cardinality is:
!
5
NM (n)
1X
NC (n) + ∆N(n)
1 12
νM (n) =
+
=
=
ζ(12i, n)(n−i)νi (n) ;
n
n
12 n
n i=1
(30)
where an inspection of Table 1 shows that, in general, the number of Tn set
classes including only distinct or repeating n-chords is different from NC (n)/n
or ∆N(n)/n, respectively.
The above results complete the calculation of ∆N, ∆N/n, within the
domain, 1 ≤ n ≤ 12, which allows the knowledge of the total number of
(distinct + repeating) n-chords, NM = NC + ∆N, and the total number of
Tn set classes, νM , which are also listed in Tables 1, 2, 3. It can be seen Tn set
classes are symmetric with respect to n = 6, within the domain, 1 ≤ n ≤ 11.
The extension of the domain to n = 0 can be made demanding symmetry
with respect to n = 12, which implies the following:
NM (12)
NM (n)
=
=1 ;
(31)
n→0
n
12
!
∆N(n)
NM (12) NC (12)
11
NM (n) NC (n)
lim+
=
= lim+
−
−
=
; (32)
n→0
n→0
n
n
n
12
12
12
lim+
as shown in Table 1.
Following a similar procedure, the number of Tn /Tn I set classes of each
cardinality can also be determined. In this view, for instance, the Tn /Tn I set
class,
{1, 2, 3, 6}, {2, 3, 6, 1}, {3, 6, 1, 2}, {6, 1, 2, 3},
{6, 3, 2, 1}, {1, 6, 3, 2}, {2, 1, 6, 3}, {3, 2, 1, 6};
is made of eight distinct 4-chords, while the Tn /Tn I set class,
{1, 5, 5, 1}, {5, 5, 1, 1}, {5, 1, 1, 5}, {1, 1, 5, 5},
{1, 5, 5, 1}, {1, 1, 5, 5}, {5, 1, 1, 5}, {5, 5, 1, 1};
is made of four distinct and four repeating 4-chords.
With regard to a n-chord of cardinality, n, let the n-chord type be defined
as ℓi11 ℓi22 ...ℓikk , where ij denotes the multiplicity of the coordinate, ℓj , 1 ≤
j ≤ k, which implies i1 + i2 + ... + ik = n. Clearly the enumeration of
distinct n-chords remains unchanged and the results valid for Tn set classes
maintain for Tn /Tn I set classes. Conversely, the number of repeating nchords is expected to grow due to larger cardinality, 2n, of Tn /Tn I set classes,
which implies additional kind of repeating n-chords with respect to Tn set
13
classes i.e. n-chords made of identical singletons, duoes, trioes, quartets,
quintets of coordinates.
In this view, let palindrome n-chords be defined as invariant with respect
to reflection, and pseudo palindrome n-chords as invariant with respect to
reflection after appropriate transposition. For instance, {1, 5, 5, 1} is palindrome while {5, 5, 1, 1} is pseudo palindrome. Palindrome and pseudo palindrome n-chords make the additional kinds of repeating n-chords, in connection with Tn /Tn I set classes. Then the number of repeating palindrome and
pseudo palindrome n-chords of each cardinality, which have not previously
considered, has to be determined.
With regard to Tn /Tn I set classes of each cardinality from 1 to 12 (with
the addition of 0), equivalence classes are made of 2n n-chords which are
related via circular permutation and reflection. The total number can be
determined along the following steps.
(i) Start from Tn set classes of each cardinality.
(ii) Separate Tn set classes exhibiting neither palindrome nor pseudo palindrome n-chords, let the total number be denoted as νN (n), from Tn set classes
exhibiting palindrome or pseudo palindrome n-chords, let the total number
be denoted as νP (n).
(iii) Determine νN (n) and νP (n).
(iv) Calculate the total number of Tn /Tn I set classes as νQ (n) = νN (n)/2 +
νP (n), according to the above considerations.
Palindrome and pseudo palindrome n-chords are necessarily made of pairs
of identical coordinates e.g., {1, 2, 3, 3, 2, 1}, for even cardinality, with the
addition of a single coordinate e.g., {1, 4, 2, 4, 1}, for odd cardinality. Then
Eq. (1) reduces to:
k1 ℓ1 + k2 ℓ2 + ... + ki ℓi = 12 ;
k1 ℓ1 + k2 ℓ2 + ... + ki ℓi + ℓi+1 = 12 ;
(33a)
(33b)
respectively, where kj , 1 ≤ j ≤ i, is the multiplicity of the coordinate, ℓj .
Let Tn set classes made of palindrome and pseudo palindrome n-chords be
denoted as Tn, P .
For n = 0, 1, n-chords remain unchanged after transposition and/or reflection. For n = 2, transposition and reflection are equivalent or, in other
words, all n-chords are palindrome or pseudo palindrome. Accordingly,
νQ (n) = νM (n), n < 3, where the number of Tn set classes, νM (n), is listed
in Table 3.
14
For n = 3, Eq. (33b) reduces to:
2ℓ1 + ℓ2 = 12 ;
n=3 ;
(34)
which has solutions as (ℓ1 , ℓ2 ) = (1, 10), (2,8), (3,6), (4,4), (5,2), implying
νP (3) = 5, νN (3) = νM (3) − νP (3) = 19 − 5 = 14; νQ (3) = νN (3)/2 + νP (3) =
14/2 + 5 = 7 + 5 = 12.
For n = 4, Eq. (33a) reduces to:
2ℓ1 + ℓ2 + ℓ3 = 12 ;
n=4 ;
2ℓ1 + 2ℓ2 = 12 ;
n=4 ;
3ℓ1 + ℓ2 = 12 ;
n=4 ;
(35a)
(35b)
(35c)
which has solutions as (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 8), (1,3,7), (1,4,6), (2,1,7), (2,3,5),
(3,1,5), (3,2,4), (4,1,3), (3,3,3); (ℓ1 , ℓ2 ) = (1, 5), (2,4); (ℓ1 , ℓ2 ) = (1, 9), (2,6);
respectively.
Case a yields one kind of Tn, P set classes, namely {x, y, x, z}. Accordingly,
νPa (4) = 1 · 9 = 9.
Case b yields two kinds of Tn, P set classes, namely {x, x, y, y}, {x, y, x, y}.
Accordingly, νPb (4) = 2 · 2 = 4.
Case c yields one kind of Tn, P set classes, namely {x, x, x, y}. Accordingly,
νPc (4) = 1 · 2 = 2.
P
Finally, νP (4) = i νPi (4) = 9 + 4 + 2 = 15; νN (4) = νM (4) − νP (4) =
43 − 15 = 28; νQ (4) = νN (4)/2 + νP (4) = 28/2 + 15 = 14 + 15 = 29.
For n = 5, Eq. (33b) reduces to:
4ℓ1 + ℓ2 = 12 ;
n=5 ;
2ℓ1 + 2ℓ2 + ℓ3 = 12 ;
n=5 ;
3ℓ1 + 2ℓ2 = 12 ;
n=5 ;
(36a)
(36b)
(36c)
which has solutions as (ℓ1 , ℓ2 ) = (1, 8), (2,4); (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 6), (1,3,4),
(1,4,2); (ℓ1 , ℓ2 ) = (2, 3); respectively.
Case a yields one kind of Tn, P set classes, namely {x, x, x, x, y}. Accordingly, νPa (5) = 1 · 2 = 2.
Case b yields two kinds of Tn, P set classes, namely {x, y, z, y, x}, {y, x, z, x, y}.
Accordingly, νPb (5) = 2 · 3 = 6.
Case c yields two kinds of Tn, P set classes, namely {x, x, x, y, y}, {x, y, x, y, x}.
Accordingly, νPc (5) = 2 · 1 = 2.
P
Finally, νP (5) = i νPi (5) = 2 + 6 + 2 = 10; νN (5) = νM (5) − νP (5) =
66 − 10 = 56; νQ (5) = νN (5)/2 + νP (5) = 56/2 + 10 = 28 + 10 = 38.
15
For n = 6, Eq. (33a) reduces to:
2ℓ1 + 2ℓ2 + 2ℓ3 = 12 ;
n=6 ;
3ℓ1 + 3ℓ2 = 12 ;
n=6 ;
3ℓ1 + 2ℓ2 + ℓ3 = 12 ;
n=6 ;
4ℓ1 + 2ℓ2 = 12 ;
n=6 ;
4ℓ1 + ℓ2 + ℓ3 = 12 ;
n=6 ;
5ℓ1 + ℓ2 = 12 ;
n=6 ;
(37a)
(37b)
(37c)
(37d)
(37e)
(37f)
which has solutions as (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 3); (ℓ1 , ℓ2 ) = (1, 3); (ℓ1 , ℓ2 , ℓ3 ) =
(1, 2, 5), (2,1,4); (ℓ1 , ℓ2 ) = (1, 4); (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 6), (1,3,5), (2,1,3); (ℓ1 , ℓ2 ) =
(1, 7), (2, 2); respectively.
Case a yields six kinds of Tn, P set classes, namely {x, y, x, z, y, z}, {x, z, x, y, z, y},
{y, x, y, z, x, z}, {x, z, y, y, z, x}, {y, x, z, z, x, y}, {z, y, x, x, y, z}. Accordingly, νPa (6) = 6 · 1 = 6.
Case b yields two kinds of Tn, P set classes, namely {x, x, x, y, y, y}, {x, y, x, y, x, y}.
Accordingly, νPb (6) = 2 · 1 = 2.
Case c yields two kinds of Tn, P set classes, namely {x, x, x, y, z, y}, {x, z, x, y, x, y}.
Accordingly, νPc (6) = 2 · 2 = 4.
Case d yields three kinds of Tn, P set classes, namely {x, x, y, y, x, x},
{x, y, x, x, y, x}, {x, x, x, y, x, y}. Accordingly, νPd (6) = 3 · 1 = 3.
Case e yields one kind of Tn, P set classes, namely {x, y, x, x, z, x}. Accordingly, νPe (6) = 1 · 3 = 3.
Case f yields one kind of Tn, P set classes, namely {x, x, x, x, x, y}. Accordingly, νPf (6) = 1 · 2 = 2.
P
Finally, νP (6) = i νPi (6) = 6 + 2 + 4 + 3 + 3 + 2 = 20; νN (6) = νM (6) −
νP (6) = 80 − 20 = 60; νQ (6) = νN (6)/2 + νP (6) = 60/2 + 20 = 30 + 20 = 50.
For n = 7, Eq. (33b) reduces to:
6ℓ1 + ℓ2 = 12 ;
n=7 ;
4ℓ1 + 2ℓ2 + ℓ3 = 12 ;
n=7 ;
5ℓ1 + 2ℓ2 = 12 ;
n=7 ;
(38a)
(38b)
(38c)
which has solutions as (ℓ1 , ℓ2 ) = (1, 6); (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 4), (1,3,2); (ℓ1 , ℓ2 ) =
(2, 1); respectively.
Case a yields one kind of Tn, P set classes, namely {x, x, x, x, x, x, y}. Accordingly, νPa (7) = 1 · 1 = 1.
Case b yields three kinds of Tn, P set classes, namely {x, x, y, z, y, x, x},
{x, y, x, z, x, y, x}, {y, x, x, z, x, x, y}. Accordingly, νPb (7) = 3 · 2 = 6.
Case c yields three kinds of Tn, P set classes, namely {y, x, x, x, x, x, y},
{x, y, , x, x, y, x}, {x, x, y, x, y, x, x}. Accordingly, νPc (7) = 3 · 1 = 3.
16
P
Finally, νP (7) = i νPi (7) = 1 + 6 + 3 = 10; νN (7) = νM (7) − νP (7) =
66 − 10 = 56; νQ (7) = νN (7)/2 + νP (7) = 56/2 + 10 = 28 + 10 = 38.
For n = 8, Eq. (33a) reduces to:
4ℓ1 + 4ℓ2 = 12 ;
n=8 ;
6ℓ1 + ℓ2 + ℓ3 = 12 ;
n=8 ;
5ℓ1 + 2ℓ2 + ℓ3 = 12 ;
n=8 ;
6ℓ1 + 2ℓ2 = 12 ;
n=8 ;
7ℓ1 + ℓ2 = 12 ;
n=8 ;
(39a)
(39b)
(39c)
(39d)
(39e)
which has solutions as (ℓ1 , ℓ2 ) = (1, 2); (ℓ1 , ℓ2 , ℓ3 ) = (1, 2, 4); (ℓ1 , ℓ2 , ℓ3 ) =
(1, 2, 3); (ℓ1 , ℓ2 ) = (1, 3); (ℓ1 , ℓ2 ) = (1, 5); respectively.
Case a yields six kinds of Tn, P set classes, namely {x, x, x, x, y, y, y, y},
{x, x, y, y, x, x, y, y}, {x, y, x, y, x, y, x, y}, {x, y, x, y, y, x, y, x}, {x, x, y, y, x, y, y, x},
{x, x, y, x, x, y, y, y}. Accordingly, νPa (8) = 6 · 1 = 6.
Case b yields one kind of Tn, P set classes, namely {x, x, x, y, x, x, x, z}.
Accordingly, νPb (8) = 1 · 1 = 1.
Case c yields three kinds of Tn, P set classes, namely {x, x, x, y, z, y, x, x},
{x, x, y, x, z, x, y, x}, {x, y, x, x, z, x, x, y}. Accordingly, νPc (8) = 3 · 1 = 3.
Case d yields four kinds of Tn, P set classes, namely {x, x, x, x, x, x, y, y},
{x, x, x, x, x, y, x, y}, {x, x, x, x, y, x, x, y}, {x, x, x, y, x, x, x, y}. Accordingly,
νPd (8) = 4 · 1 = 4.
Case e yields one kind of Tn, P set classes, namely {x, x, x, x, x, x, x, y}.
Accordingly, νPe (8) = 1 · 1 = 1.
P
Finally, νP (8) = i νPi (8) = 6+1+3+4+1 = 15; νN (8) = νM (8)−νP (8) =
43 − 15 = 28; νQ (8) = νN (8)/2 + νP (8) = 28/2 + 15 = 14 + 15 = 29.
For n = 9, Eq. (33b) reduces to:
8ℓ1 + ℓ2 = 12 ;
6ℓ1 + 3ℓ2 = 12 ;
n=9 ;
n=9 ;
(40a)
(40b)
which has solutions as (ℓ1 , ℓ2 ) = (1, 4); (ℓ1 , ℓ2 ) = (1, 2); respectively.
Case a yields one kind of Tn, P set classes, namely {x, x, x, x, x, x, x, x, y}.
Accordingly, νPa (9) = 1 · 1 = 1.
Case b yields four kinds of Tn, P set classes, namely {x, x, x, y, y, y, x, x, x},
{x, x, y, x, y, x, y, x, x}, {x, y, x, x, y, x, x, y, x}, {y, x, x, x, y, x, x, x, y}. Accordingly, νPb (9) = 4 · 1 = 4.
P
Finally, νP (9) = i νPi (9) = 1 + 4 = 5; νN (9) = νM (9) − νP (9) = 19 − 5 =
14; νQ (9) = νN (9)/2 + νP (9) = 14/2 + 5 = 7 + 5 = 12.
For n = 10, Eq. (33a) reduces to:
8ℓ1 + 2ℓ2 = 12 ;
9ℓ1 + ℓ2 = 12 ;
17
n = 10 ;
n = 10 ;
(41a)
(41b)
Table 4: Number of Tn set classes, νM (n), palindrome and pseudo palindrome Tn set classes, νP (n), neither palindrome nor pseudo palindrome
Tn set classes, νN (n) = νM (n) − νP (n), and Tn /Tn I set classes, νQ (n) =
νN (n)/2 + νP (n), involving n-chords of each cardinality, n, 0 ≤ n ≤ 12. See
text for further details.
n
νM
νP
νN
νQ
0
1
1
0
1
1
1
1
0
1
2 3 4 5
6 19 43 66
6 5 15 10
0 14 28 56
6 12 29 38
6
80
20
60
50
7
66
10
56
38
8 9 10
43 19 6
15 5 6
28 14 0
29 12 6
11 12
1 1
1 1
0 0
1 1
which has solutions as (ℓ1 , ℓ2 ) = (1, 2); (ℓ1 , ℓ2 ) = (1, 4); respectively.
Case a yields five kinds of Tn, P set classes, namely {x, x, x, x, x, x, x, x, y, y},
{x, x, x, x, x, x, x, y, x, y}, {x, x, x, x, x, x, y, x, x, y}, {x, x, x, x, x, y, x, x, x, y},
{x, x, x, x, y, x, x, x, x, y}. Accordingly, νPa (10) = 5 · 1 = 5.
Case b yields one kind of Tn, P set classes, namely {x, x, x, x, x, x, x, x, x, y}.
Accordingly, νPb (10) = 1 · 1 = 1.
P
Finally, νP (10) = i νPi (10) = 5 + 1 = 6; νN (10) = νM (10) − νP (10) =
6 − 6 = 0; νQ (10) = νN (10)/2 + νP (10) = 0/2 + 6 = 0 + 6 = 6.
For n = 11, Eq. (33b) reduces to:
10ℓ1 + ℓ2 = 12 ;
n = 11 ;
(42)
which has solutions as (ℓ1 , ℓ2 ) = (1, 2), yielding one kind of Tn, P set classes,
namely {x, x, x, x, x, x, x, x, x, x, y}. Accordingly, νP (11) = 1·1 = 1; νN (11) =
νM (11)−νP (11) = 1−1 = 0; νQ (11) = νN (11)/2+νP (11) = 0/2+1 = 0+1 = 1.
For n = 12, Eq. (33a) reduces to:
12ℓ1 = 12 ;
n = 12 ;
(43)
which has solutions as ℓ1 = 1, yielding one kind of Tn, P set classes, namely
{x, x, x, x, x, x, x, x, x, x, x, x}. Accordingly, νP (12) = 1 · 1 = 1; νN (12) =
νM (12)−νP (12) = 1−1 = 0; νQ (12) = νN (12)/2+νP (12) = 0/2+1 = 0+1 = 1.
The total number of Tn set classes, νM (n) (listed in Table 3 and repeated
for better comparison), palindrome and pseudo palindrome Tn set classes,
νP (n), neither palindrome nor pseudo palindrome Tn set classes, νN (n) =
νM (n) − νP (n), and Tn /Tn I set classes, νQ (n) = νN (n)/2 + νP (n), are listed
in Table 4.
An inspection of Table 4 shows a symmetry with respect to n = 6. This
is why complementation gives a one to one correspondence between Tn set
18
classes of cardinality, n and 12−n, respectively, which is preserved for Tn /Tn I
set classes e.g., [14] Chap. 9 §9.14. The same kind of symmetry is also implicit
in the binomial formula, expressed by Eq. (6).
3
Geometrical interpretation
With regard to an Euclidean n-space, ℜn , and a Cartesian orthogonal reference frame, (O x1 x2 , ... xn ), the extension of the boundary condition, expressed by Eq. (1), to real numbers, reads:
x1 + x2 + ... + xn = 12 ;
1 ≤ n ≤ 12 ;
(44)
which represents a (n−1)-plane intersecting the coordinate axes at the points,
Vk ≡ (12δ1k , 12δ2k , ..., 12δnk ), 1 ≤ k ≤ n, where δik is the Kronecker symbol.
The following properties can be established: (i) the (n − 1)-plane, expressed by Eq. (44), is normal to the n-sector (n = 2, bisector; n = 3, trisector; and so on) of the positive 2n -ant; (ii) the region of (n−1)-plane, bounded
by the positive 2n -ant, is a regular, inclined n-hedron, Ψn12 , of vertexes, Vk ,
1 ≤ k ≤ n; (iii) special cases are Ψ112 , regular vertex; Ψ212 , regualr side;
Ψ312 , regular triangle; Ψ412 , regular tetrahedron; (iv) the orthocentre of Ψn12 ,
Hn ≡ (12/n, 12/n, ..., 12/n), is the intersection between the (n − 1)-plane and
the n-sector of the positive 2n -ant. For further details, an interested reader
is addressed to Appendix C.
In general, Ψn12 may be divided into n congruent n-hedrons, Ψn12,i , 1 ≤
i ≤ n, by joining the vertexes with the orthocentre. More specifically, the
orthocentre is the common vertex while the remaining (n − 1) vertexes lie on
a (n − 2) hyperface, or (n − 2)-face, of Ψn12 .
The special case, Ψ312 , is represented in Fig. 2 and related n-chords are
shown as coordinates of positive integer points, satisfying the boundary condition expressed by Eq. (1), in Fig. 3.
According to the above considerations, n-chords are represented as coordinates of positive integer points within Ψn12 i.e. satisfying the boundary
condition expressed by Eq. (1). Tn and Tn /Tn I set classes contain n and
n+n = 2n points of the kind considered, respectively, which implies (i) points
with distinct coordinates, belonging to the same Tn or Tn /Tn I set class, are
similarly placed within different Ψn12,i , 1 ≤ i ≤ n; (ii) points with identical soloes, duoes, trioes, quartets, and quintets of coordinates are placed
on (n − 2)-faces between different Ψn12,i , i = i1 , i2 , 1 ≤ i1 < i2 ≤ n; (iii)
positive integer points (taking into due account the multiciplity of repeating n-chords) are equally partitioned among Ψn12,i , 1 ≤ i ≤ n, in number
19
of νM (n) = NM (n)/n via Eq. (30). For complementary Euclidean n-spaces,
ℜn ↔ ℜ12−n , νM (n) = νM (12 − n) as shown in Table 3.
In general, Ψn12,i , 1 ≤ i ≤ n, have basis coinciding with the ith (n−2)-face
of Ψn12 , which implies (n − 1) vertexes in common, with the inclusion of the
orthocentre of Ψn12 . Accordingly, the coordinates of a generic positive integer
point, Pi ≡ (ℓ1 , ℓ2 , ..., ℓn ), belonging to Ψn12,i , 1 ≤ i ≤ n, have necessarily to
satisfy the conditions:
1 ≤ ℓi ≤
12
;
n
1 ≤ ℓj ≤ 12−(n−1) ;
1 ≤ j ≤ 12 ;
j 6= i ; (45)
where the vertex of Ψn12 , placed on the coordinate axis, xi , does not belong to
the congruent n-hedron under consideration. It is apparent circular permutation of coordinates makes the points, Pi , be similarly placed within different
Ψn12,i , 1 ≤ i ≤ n, until the initial configuration is attained.
The extension of the symmetry, outlined in Table 1, to the special case,
n = 12, implies Euclidean 0-spaces, ℜ0 , to be taken into consideration.
Accordingly, Ψ012 would lie outside ℜ0 , on a coordinate axis at a distance,
x0 = 12, from the origin which, in turn, coincides with ℜ0 . Then the boundary condition, expressed by Eq. (1), is satisfied.
At this stage, a nontrivial question is if n-chords of each cardinality can be
enumerated within the framework of the geometrical interpretation outlined
above. Let n-chords exhibiting distinct coordinates be first considered.
For n = 0, the Euclidean 0-space, ℜ0 , reduces to the origin and Eq. (1)
still holds but with regard to a point of coordinate, ℓ0 = 12, outside ℜ0 .
Accordingly, NC (0) = 0 as listed in Table 1.
For n = 1, the Euclidean 1-space, ℜ1 , reduces to the real axis, where
Eq. (1) is satisfied on the point of coordinate, ℓ1 = 12. Accordingly, NC (1) =
1, as listed in Table 1.
For n = 2, the Euclidean 2-space, ℜ2 , reduces to a plane, where Eq. (1) is
satisfied on the points of coordinates, (ℓ1 , 12 − ℓ1 ), 1 ≤ ℓ1 ≤ 11. The above
mentioned points are displaced on a regular inclined 2-hedron (regular side),
2
ψ12
, of vertexes, Vi′ ≡ (1 + 10δ1i , 1 + 10δ2i ), 1 ≤ i ≤ 2, in number of eleven.
Accordingly, the number of distinct 2-chords must be counted along a series
of superimposed 1-hedrons (regular vertexes) through the second dimension.
The result is NC (2) = 11, as listed in Table 1.
For n = 3, the Euclidean 3-space, ℜ3 , is an ordinary space, where Eq. (1)
is satisfied on the points of coordinates, (ℓ1 , ℓ2 , 12 − ℓ1 − ℓ2 ), 1 ≤ ℓk ≤ 10,
k = 1, 2. As shown in Fig. 3, the above mentioned points are displaced on a
3
regular inclined 3-hedron (regular triangle), ψ12
, of vertexes, Vi′ ≡ (1+9δ1i , 1+
9δ2i , 1 + 9δ3i ), 1 ≤ i ≤ 3, in number of ten on each 1-face (regular side) and
scaled by one passing to the next related 1-hedron up to the opposite vertex.
20
Accordingly, the number of distinct 3-chords must be counted along a series
of superimposed 2-hedrons through the third dimension. The result is:
NC (3) =
10
X
11 · 10
= 55 ; (46)
2
k = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 =
k=1
as listed in Table 1.
For n = 4, Eq. (1) is satisfied in ℜ4 on the points of coordinates, (ℓ1 , ℓ2 , ℓ3 ,
12 − ℓ1 − ℓ2 − ℓ3 ), 1 ≤ ℓk ≤ 9, k = 1, 2, 3. The above mentioned points
4
are displaced on a regular inclined 4-hedron (regular tetrahedron), ψ12
, of
′
vertexes, Vi ≡ (1+8δ1i , 1+8δ2i , 1+8δ3i , 1+8δ4i ), 1 ≤ i ≤ 4, in number of nine
on each 1-face, yielding 10 · 9/2 = 45 points on each 2-face. Accordingly, the
number of distinct 4-chords must be counted along a series of superimposed
3-hedrons through the fourth dimension. The result is:
NC (4) =
9
X
k+
k=1
8
X
k+
k=1
7
X
k+
k=1
6
X
k+
k=1
5
X
4
X
k+
k=1
3
X
k+
k=1
k=1
k+
2
X
k+
k=1
=5·9+4·9+4·7+3·7+3·5+2·5+2·3+1·3+1·1
= 92 + 72 + 52 + 32 + 12 = 165 ;
1
X
k
k=1
(47)
as listed in Table 1.
For generic n, Eq. (1) is satisfied in ℜn on the points of coordinates,
(ℓ1 , ..., ℓn−1, 12 − ℓ1 − ... − ℓn−1 ), 1 ≤ ℓk ≤ 12 − n + 1, k = 1, 2, ..., n − 1. The
n
above mentioned points are displaced on a regular inclined n-hedron, ψ12
, of
′
vertexes, Vi ≡ [1 + (12 − n)δ1i , 1 + (12 − n)δ2i , ..., 1 + (12 − n)δni ], 1 ≤ i ≤ n,
in number of 12 − n + 1 on each 1-face, yielding (12 − n + 2)(12 − n + 1)/2
points on each 2-face. Accordingly, the number of distinct n-chords must
be counted along a series of superimposed (n − 1)-hedrons through the nth
dimension. The result is:
(n−2)
(n−2)
NC (n) = NC τ12−n+1 + NC τ12−n
(n−2)
where NC τk
(n−2)
+ ... + NC τ1
;
n ≥ 2 ; (48)
represents the number of distinct n-chords within the
(n−2)
(n − 1)-hedron of the series, τk
, 1 ≤ k ≤ 12 − n + 1, including on each
1-face k positive integer points which satisfy Eq. (1). More specifically, the
generic term on the right-hand side of Eq. (48) can be expressed as:
(n−2)
NC τk
(n−3)
where NC τk+1
(n−2)
= NC τk+1
(n−3)
− NC τk+1
(n−2)
is the counterpart of NC τk+1
;
n≥2 ;
(49)
with regard to the (n−2)-
(n−2)
hedron of the related series in ℜn−1 , and NC τ12−n+2 = NC (n − 1). Then
21
each term on the right-hand side of Eq. (48) can be determined via Eq. (49),
provided its counterpart in ℜn−1 is known.
The particularization of Eq. (49) to n = 5 via (47) yields:
(3)
NC τ8
(3)
NC τ7
(3)
NC τ6
(3)
NC τ5
(3)
NC τ4
(3)
NC τ3
NC
(3)
τ2
(3)
NC τ1
(2)
= NC (4) − NC τ9
(3)
= NC τ8
(3)
= NC τ7
(3)
= NC τ6
(3)
= NC τ5
(3)
= NC τ4
= NC
(3)
τ3
(3)
= NC τ2
= 165 − 45 = 120 ;
(2)
− NC τ8
(2)
− NC τ7
(2)
− NC τ6
(2)
− NC τ5
(2)
− NC τ4
− NC
(2)
τ3
(2)
− NC τ2
(50a)
= 120 − 36 = 84 ;
(50b)
= 84 − 28 = 56 ;
(50c)
= 56 − 21 = 35 ;
(50d)
= 35 − 15 = 20 ;
(50e)
= 20 − 10 = 10 ;
(50f)
= 10 − 6 = 4 ;
(50g)
=4−3=1 ;
(50h)
and the particularization of Eq. (48) to n = 5 via (50) yields the number of
distinct 5-chords as:
NC (5) = 120 + 84 + 56 + 35 + 20 + 10 + 4 + 1 = 330 ;
(51)
in accordance with Table 1.
The particularization of Eq. (49) to n = 6 via (50)-(51) yields:
(4)
NC τ7
(4)
NC τ6
(4)
NC τ5
(4)
NC τ4
(4)
NC τ3
(4)
NC τ2
(4)
NC τ1
(3)
= NC (5) − NC τ8
(4)
= NC τ7
(4)
= NC τ6
(4)
= NC τ5
(4)
= NC τ4
(4)
= NC τ3
(4)
= NC τ2
= 330 − 120 = 210 ;
(3)
− NC τ7
(3)
− NC τ6
(3)
− NC τ5
(3)
− NC τ4
(3)
− NC τ3
(3)
− NC τ2
(52a)
= 210 − 84 = 126 ;
(52b)
= 126 − 56 = 70 ;
(52c)
= 70 − 35 = 35 ;
(52d)
= 35 − 20 = 15 ;
(52e)
= 15 − 10 = 5 ;
(52f)
= 5−4 =1 ;
(52g)
and the particularization of Eq. (48) to n = 6 via (52) yields the number of
distinct 6-chords as:
NC (6) = 210 + 126 + 70 + 35 + 15 + 5 + 1 = 462 ;
in accordance with Table 1.
22
(53)
The particularization of Eq. (49) to n = 7 via (52)-(53) yields:
(5)
NC τ6
(5)
NC τ5
(5)
NC τ4
(5)
NC τ3
(5)
NC τ2
(5)
NC τ1
(4)
= NC (6) − NC τ7
(5)
= NC τ6
(5)
= NC τ5
(5)
= NC τ4
(5)
= NC τ3
(5)
= NC τ2
= 462 − 210 = 252 ;
(4)
− NC τ6
(4)
− NC τ5
(4)
− NC τ4
(4)
− NC τ3
(4)
− NC τ2
(54a)
= 252 − 126 = 126 ;
(54b)
= 126 − 70 = 56 ;
(54c)
= 56 − 35 = 21 ;
(54d)
= 21 − 15 = 6 ;
(54e)
=6−5=1 ;
(54f)
and the particularization of Eq. (48) to n = 7 via (54) yields the number of
distinct 7-chords as:
NC (7) = 252 + 126 + 56 + 21 + 6 + 1 = 462 ;
(55)
in accordance with Table 1.
The particularization of Eq. (49) to n = 8 via (54)-(55) yields:
(6)
NC τ5
(6)
NC τ4
(6)
NC τ3
(6)
NC τ2
(6)
NC τ1
(5)
= NC (7) − NC τ6
(6)
= NC τ5
(6)
= NC τ4
(6)
= NC τ3
(6)
= NC τ2
= 462 − 252 = 210 ;
(5)
− NC τ5
(5)
− NC τ4
(5)
− NC τ3
(5)
− NC τ2
(56a)
= 210 − 126 = 84 ;
(56b)
= 84 − 56 = 28 ;
(56c)
= 28 − 21 = 7 ;
(56d)
= 7−6 =1 ;
(56e)
and the particularization of Eq. (48) to n = 8 via (56) yields the number of
distinct 8-chords as:
NC (8) = 210 + 84 + 28 + 7 + 1 = 330 ;
(57)
in accordance with Table 1.
The particularization of Eq. (49) to n = 9 via (56)-(57) yields:
(7)
NC τ4
(7)
NC τ3
(7)
NC τ2
(7)
NC τ1
(6)
= NC (8) − NC τ5
(7)
= NC τ4
(7)
= NC τ3
(7)
= NC τ2
= 330 − 210 = 120 ;
(6)
− NC τ4
(6)
− NC τ3
(6)
− NC τ2
23
(58a)
= 120 − 84 = 36 ;
(58b)
= 36 − 28 = 8 ;
(58c)
=8−7=1 ;
(58d)
and the particularization of Eq. (48) to n = 9 via (58) yields the number of
distinct 9-chords as:
NC (9) = 120 + 36 + 8 + 1 = 165 ;
(59)
in accordance with Table 1.
The particularization of Eq. (49) to n = 10 via (58)-(59) yields:
(8)
NC τ3
NC
(8)
τ2
(8)
NC τ1
(7)
= NC (9) − NC τ4
= NC
(8)
τ3
(8)
= NC τ2
− NC
= 165 − 120 = 45 ;
(60a)
= 45 − 36 = 9 ;
(60b)
=9−8=1 ;
(60c)
(7)
τ3
(7)
− NC τ2
and the particularization of Eq. (48) to n = 10 via (60) yields the number of
distinct 10-chords as:
NC (10) = 45 + 9 + 1 = 55 ;
(61)
in accordance with Table 1.
The particularization of Eq. (49) to n = 11 via (60)-(61) yields:
(9)
NC τ2
(9)
NC τ1
(8)
= NC (10) − NC τ3
(9)
= NC τ2
= 55 − 45 = 10 ;
(62a)
= 10 − 9 = 1 ;
(62b)
(8)
− NC τ2
and the particularization of Eq. (48) to n = 11 via (62) yields the number of
distinct 11-chords as:
NC (11) = 10 + 1 = 11 ;
(63)
in accordance with Table 1.
The particularization of Eq. (49) to n = 12 via (62)-(63) yields:
(10)
NC τ1
(9)
= NC (11) − NC τ2
= 11 − 10 = 1 ;
(64)
and the particularization of Eq. (48) to n = 12 via (64) yields the number of
distinct 12-chords as:
NC (12) = 1 ;
(65)
in accordance with Table 1.
In summary, Eqs. (46)-(65) disclose Eq. (5) can be inferred via geometric
as well as algebraic considerations, but the former alternative deserves further
attention. To this respect, let regular, inclined n-hedrons, Ψn12 , of vertexes,
24
Vi ≡ (12δ1i , 12δ2i , ..., 12δni ), 1 ≤ i ≤ n, be considered as special cases of
lattice polytopes in ℜn e.g., [22].
In this view, Eq. (5) yields the number of lattice points (i.e. points of
integer coordinates) within Ψn12 , while the total number reads:
NV (n) = NC (n) + NS (n) ;
(66)
where NS (n) is the number of lattice points on the (n − 2)-surface of Ψn12 ,
which necessarily exhibit at least one null coordinate. In the following, NS (n)
and NV (n) shall be determined for each cardinality, keeping in mind values
of NC (n) are listed in Table 1.
For n = 0, no lattice polytope exists in the sense Ψ012 lies outside ℜ0 .
Accordingly, NS (0) = 0, NV (0) = 0.
For n = 1, Ψ112 reduces to a regular vertex, Vi ≡ (12δ1i ). Accordingly,
NS (1) = 0, NV (1) = NC (1) = 1.
For n = 2, Ψ212 reduces to a regular side of extremes, Vi ≡ (12δ1i , 12δ2i ).
Accordingly, NS (2) = 1 · 2 = 2, NV (2) = 11 + 2 = 13.
For n = 3, Ψ312 reduces to a regular triangle of vertexes, Vi ≡ (12δ1i , 12δ2i , 12δ3i ),
as shown in Fig. 2. Lattice points exhibiting at least one null coordinate
lie on the principal planes, in number of 12 provided vertexes (exhibiting
two null coordinates) are partitioned one per principal plane. Accordingly,
NS (2) = 3 · 12 = 36, NV (2) = 55 + 36 = 91.
In general, passing from ℜn−1 to ℜn , the number of lattice points within
n−1
n−2
Ψ12
has to be incremented by the number of lattice points within Ψ12
for including points exhibiting at least one null coordinate, which makes the
number of lattice points on a (n−2)-face of Ψn12 i.e. 1/n the number of lattice
points on the (n−2)-surface of Ψn12 . Accordingly, the number of lattice points
on the (n − 2)-surface of Ψn12 reads:
NS (n) = n[NC (n − 1) + NC (n − 2)] ;
n>1 ;
(67)
which, for n = 2, 3, reproduces the above results.
Using Eqs. (66)-(67) for each cardinality yields:
NS (2) = 2(1 + 0) = 2 ;
NV (2) = 11 + 2 = 13 ;
NS (3) = 3(11 + 1) = 36 ;
NV (3) = 55 + 36 = 91 ;
NS (4) = 4(55 + 11) = 264 ;
NV (4) = 165 + 264 = 429 ;
NS (5) = 5(165 + 55) = 1100 ;
NV (5) = 330 + 1100 = 1430 ;
NS (6) = 6(330 + 165) = 2970 ;
NV (6) = 462 + 2970 = 3432 ;
NS (7) = 7(462 + 330) = 5544 ;
NV (7) = 462 + 5544 = 6006 ;
NS (8) = 8(462 + 462) = 7392 ;
NV (8) = 330 + 7392 = 7722 ;
25
(68a)
(68b)
(68c)
(68d)
(68e)
(68f)
(68g)
Table 5: Number of lattice points within the (n − 1)-volume, NC , and on the
(n − 2)-surface, NS , of Ψn12 polytopes, and total amount, NV = NC + NS . See
text for further details.
n
NC
NS
NV
0
0
0
0
1
1
0
1
2 3
4
5
6
7
8
9
10
11 12
11 55 165 330 462 462 330 165
55
11
1
2 36 264 1110 2970 5944 7392 7128 4950 2420 792
13 91 429 1430 3432 6006 7722 7293 5005 2431 793
NS (9) = 9(330 + 462) = 7128 ;
NV (9) = 165 + 7128 = 7293 ; (68h)
NS (10) = 10(165 + 330) = 4950 ;
NV (10) = 55 + 4950 = 5005 ; (68i)
NS (11) = 11(55 + 165) = 2420 ;
NV (11) = 11 + 2420 = 2431 ; (68j)
NS (12) = 12(11 + 55) = 792 ;
NV (12) = 1 + 792 = 793 ;
(68k)
which are listed in Table 5 together with NC (n), the last repeated for better
comparison.
At this stage, what remains to be done is the enumeration of repeating
n-chords within the framework of the current geometrical interpretation. To
this respect, it is worth remembering Ψn12 can be partitioned into n congruent
Ψn12,i , 1 ≤ i ≤ n, where different n-chords within an assigned Tn or Tn /Tn I
set class relate to points which are similarly placed within different Ψn12,i .
In this view, repeating n-chords necessarily lie on the boundary between
different Ψn12,i i.e. (n − 2)-faces, and must be counted twice, three times,
or more, according if points under consideration are common to two, three,
or more Ψn12,i . Then repeating n-chords are expected to relate to some kind
of symmetric points, lying on the straight lines joining the orthocentres of
(j − 1)-faces, 1 ≤ j ≤ n − 2.
For n = 0, Ψ012 lies outside ℜ0 , which implies no repeating 0-chords.
For n = 1, Ψ112 = Ψ112,1 , which implies no repeating 1-chords.
For n = 2, Ψ212,1 and Ψ212,2 have in common the orthocentre of Ψ212 , H2 ≡
(6, 6), which has to be counted twice for being equally partitioned among
Ψ212,1 and Ψ212,2 , for a total of repeating 1. In summary, the number of
repeating 2-chords is 1, as shown in Table 2.
For n = 3, Ψ312,1 , Ψ312,2 , Ψ312,3 , have in common the orthocentre of Ψ312 ,
H3 ≡ (4, 4, 4), which has to be counted three times for being equally partitioned among Ψ312,1 , Ψ312,2 , Ψ312,3 , for a total of repeating 2. In summary, the
number of repeating 3-chords is 2, as shown in Table 2.
For n = 4, Ψ412,i , 1 ≤ i ≤ 4, have in common the orthocentre of Ψ412 , H4 ≡
26
(3, 3, 3, 3), which has to be counted four times for being equally partitioned
among Ψ412,i , 1 ≤ i ≤ 4, for a total of repeating 3. In addition, pairs of 2-faces
(regular triangles) have in common orthocentres of 1-faces (regular sides) on
the boundary, H4,2 ≡ [ℓ+k(δ1i +δ1j ), ℓ+k(δ2i +δ2j ), ..., ℓ+k(δ4i +δ4j )], ℓ = 1, 2,
k = 4, 2. The number of 1-face orthocentres equals 6 in any case, where
4 are equally partitioned among Ψ412,i , while the remaining 2, {x, y, x, y},
{y, x, y, x}, must necessarily be counted twice for a total of repeating 2+2=4.
In summary, the number of repeating 4-chords is 3+4=7, as shown in Table
2.
For n = 5, the orthocentre of Ψ512 , is not an integer point. On the other
hand, (j − 1)-faces of Ψ512 , 0 < j < 4, occur in multiples of 5 as shown in
Table 7, which implies related orthocentres are equally partitioned among
Ψ512,i , 1 ≤ i ≤ 5, and repeating 5-chords do not take place, according to
Table 2.
For n = 6, Ψ612,i , 1 ≤ i ≤ 6, have in common the orthocentre of Ψ612 ,
H6 ≡ (2, 2, 2, 2, 2, 2), which has to be counted six times for being equally
partitioned among Ψ612,i , 1 ≤ i ≤ 6, as shown in Table 2. In addition, pairs
of 2-faces (regular triangles) have in common orthocentres of 1-faces (regular
sides) on the boundary, H6,2 ≡ [ℓ+k(δ1i +δ1j ), ℓ+k(δ2i +δ2j ), ..., ℓ+k(δ6i +δ6j )],
ℓ = 1, k = 3. The number of 1-face orthocentres equals 15, where 12
are equally partitioned among Ψ612,i , while the remaining 3, {x, x, y, x, x, y},
{x, y, x, x, y, x}, {y, x, x, y, x, x}, must necessarily be counted twice for a total
of repeating 1+1+1 = 3. Furthermore, pairs of 3-faces (regular tetrahedrons)
have in common orthocentres of 2-faces (regular triangles) on the boundary,
H6,3 ≡ [ℓ + k(δ1i + δ1j + δ1ℓ ), ℓ + k(δ2i + δ2j + δ2ℓ ), ..., ℓ + k(δ6i + δ6j + δ6ℓ )], ℓ = 1,
k = 2. The number of 2-face orthocentres equals 20, where 18 are equally partitioned among Ψ612,i , while the remaining 2, {x, y, x, y, x, y}, {y, x, y, x, y, x},
must necessarily be counted three times for a total of repeating 2 + 2 = 4.
Finally, triads of 3-faces (regular tetrahedrons) have in common symmetric
points with respect to the above mentioned orthocentres, exhibiting identical pairs of coordinates e.g., {x, x,
y,
y, z, z}, z = (x + y)/2. The total
6
number of pairs within a 6-tuple is 2 = 15, and the total number of pairs
within the remaining 4-tuple is 42 = 6, while the remaining 2-tuple is fixed.
Accordingly, the total number of 6-tuples with pairs of identical coordinates
equals 15 · 6 = 90, where 84 are equally partitioned among Ψ612,i , while the remaining 6, {x, y, z, x, y, z}, {y, z, x, y, z, x}, {z, x, y, z, x, y}; {x, z, y, x, z, y},
{z, y, x, z, y, x}, {y, x, z, y, x, z}; must necessarily be counted twice for a total
of repeating (1 + 1 + 1) + (1 + 1 + 1) = 3 + 3 = 6. In summary, the number
of repeating n-chords is 5+3+4+6=18, as shown in Table 2.
For n = 7, the orthocentre of Ψ712 , is not an integer point. On the other
27
hand, (j − 1)-faces of Ψ712 , 0 < j < 6, occur in multiples of 7 as shown in
Table 7, which implies related orthocentres are equally partitioned among
Ψ712,i , 1 ≤ i ≤ 7, and repeating 7-chords do not take place, according to
Table 2.
For n = 8, the orthocentre of Ψ812 , is not an integer point. On the
other hand, pairs of 2-faces (regular triangles) have in common orthocentres of 1-faces (regular sides) on the boundary, H8,2 ≡ [ℓ + k(δ1i + δ1j ), ℓ +
k(δ2i + δ2j ), ..., ℓ + k(δ8i + δ8j )], ℓ = 1, k = 2. The number of 1-face orthocentres equals 28, where 24 are equally partitioned among Ψ812,i , while the
remaining 4, {x, x, x, y, x, x, x, y}, {x, x, y, x, x, x, y, x}, {x, y, x, x, x, y, x, x},
{y, x, x, x, y, x, x, x}, must necessarily be counted twice for a total of repeating 1+1+1+1 = 4. In addition, pairs of 4-faces have in common orthocentres
of 3-faces (regular tetrahedrons) on the boundary, H8,4 ≡ [ℓ + k(δ1i1 + δ1i2 +
δ1i3 + δ1i4 ), ℓ + k(δ2i1 + δ2i2 + δ2i3 + δ2i4 ), ..., ℓ + k(δ8i1 + δ8i2 + δ8i3 + δ8i4 )],
ℓ = 1, k = 2. The number of 3-face orthocentres equals 70, where 64 are
equally partitioned among Ψ812,i , while the remaining 6, {x, y, x, y, x, y, x, y},
{y, x, y, x, y, x, y, x}; {x, x, y, y, x, x, y, y}, {x, y, y, x, x, y, y, x}, {y, y, x, x, y, y, x, x},
{y, x, x, y, y, x, x, y}; must necessarily be counted four times in the former
case and twice in the latter, for a total of repeating 3+3=6 and 1+1+1+1=4,
respectively, yielding 6+4=10. In summary, the number of repeating 8-chords
is 4+10=14, as shown in Table 2.
For n = 9, the orthocentre of Ψ912 , is not an integer point. On the other
hand, pairs of 3-faces (regular tetrahedrons) have in common orthocentres of
2-faces (regular triangles) on the boundary, H9,3 ≡ [ℓ + k(δ1i + δ1j + δ1m ), ℓ +
k(δ2i + δ2j + δ2m ), ..., ℓ + k(δ9i + δ9j + δ9m )], ℓ = 1, k = 2. The number
of 2-face orthocentres equals 84, where 81 are equally partitioned among
Ψ912,i , while the remaining 3, {x, x, y, x, x, y, x, x, y}, {x, y, x, x, y, x, x, y, x},
{y, x, x, y, x, x, y, x, x}, must necessarily be counted three times for a total of
repeating 2+2+2=6. In summary, the number of repeating 9-chords equals
6, as shown in Table 2.
For n = 10, the orthocentre of Ψ10
12 , is not an integer point. On the other
hand, pairs of 2-faces (regular triangles) have in common orthocentres of 1faces (regular sides) on the boundary, H10,2 ≡ [ℓ + k(δ1i + δ1j ), ℓ + k(δ2i +
δ2j ), ..., ℓ + k(δ10,i + δ10,j )], ℓ = 1, k = 1. The number of 1-face orthocentres
equals 45, where 40 are equally partitioned among Ψ10
12,i , while the remaining
5, {x, x, x, x, y, x, x, x, x, y}, {x, x, x, y, x, x, x, x, y, x}, {x, x, y, x, x, x, x, y, x, x},
{x, y, x, x, x, x, y, x, x, x}, {y, x, x, x, x, y, x, x, x, x}, must necessarily be counted
twice for a total of repeating 1+1+1+1+1=5. In summary, the number of
repeating 10-chords equals 5, as shown in Table 2.
For n = 11, the orthocentre of Ψ11
12 , is not an integer point. On the other
hand, (j − 1)-faces of Ψ11
,
0
<
j
<
10, occur in multiples of 11 as shown
12
28
in Table 7, which implies related orthocentres are equally partitioned among
Ψ11
12,i , 1 ≤ i ≤ 11, and repeating 11-chords do not take place, according to
Table 2.
12
For n = 12, Ψ12
12,i , 1 ≤ i ≤ 12, have in common the orthocentre of Ψ12 ,
H12 ≡ (1, 1, ..., 1), which has to be counted twelve times for being equally
partitioned among Ψ12
12,i , 1 ≤ i ≤ 12. Keeping in mind no additional positive
integer point belongs to Ψ12
12 , the number of repeating 12-chords equals 11,
as shown in Table 2.
The above results disclose both distinct and repeating n-chords and Tn or
Tn /Tn I set classes may be enumerated in the light of a geometrical interpretation, reproducing results inferred via an algebraic approach. To this respect,
it is worth noticing pairwise Tn and Tn I set classes, made of repeating but
neither palindrome nor pseudo palindrome n-chords, are different, which implies a 2 : 1 correspondence between Tn and Tn /Tn I set classes. Conversely,
pairwise Tn and Tn I set classes, made of repeating palindrome or pseudo
palindrome n-chords, are coinciding, which implies a 1 : 1 correspondence
between Tn and Tn /Tn I set classes.
For further details on Ψn12 n-hedrons, an interested reader is addressed to
Appendix C.
4
Discussion
Concerning musical intervals in multiples of semitones under 12-note equal
temperament, counting pitch-class sets with respect to an assigned equivalence relation lies between two extreme cases, namely brute-force listing and
elegant (but esoteric) enumeration procedure from group theory. The method
exploited in Section 2 implies, on one hand, determination of repeating and
palindrome or pseudo palindrome n-chords by use, on the other hand, of
a less elegant (but less esoteric) enumeration procedure. More specifically,
n-chords of cardinality, n, are related to positive integer points in ℜn , belonging to a regular inclined n-hedron, Ψn12 , where vertexes are placed on the
coordinate axes of a Cartesian orthogonal reference frame, (Ox1 x2 ... xn ), at
a distance, xi = 12, 1 ≤ i ≤ 12, from the origin.
The method is aimed to help musicians with basic knowledge of algebra,
combinatorics and group theory, but desirous of following a line of thought to
get results. This is why, say, the binomial theorem has been inferred instead
of being directly used. The procedure can be generalized to musical intervals
in multiples of semitones under L-note (instead of 12-note) equal temperament and equivalence relations other than Tn , Tn /Tn I, even if expected to
be more cumbersome. Of course, group theory remains the more powerful
29
tool to this respect, and the current approach should, ultimately, predispose
the reader towards further deepening on this subject.
According to Eqs. (5) and (28), the number of Tn set classes of cardinality,
n, reads:
NM (n)
NC (n) + ∆N(n)
=
n !
n
12
1X
1 12
+
=
ζ(12i, n)(n − i)νi (n) ;
12 n
n i=1
νM (n) =
(69)
where ζ(m1, m2 ) is defined by Eq. (19) and νi (n) is listed in Table 3, keeping
in mind i > n and/or n > 5 implies νi (n) = 0.
Within the framework of group theory, the result is e.g., [15] §72:
12
12/j
1 X
ζ(12, j)ζ(n, j)φ(j)
νM (n) =
n/j
12 j=1
!
;
(70)
where φ(k) is the Euler phi function e.g., [14] Chap. 9 §9.7 and the product,
ζ(12, j)ζ(n, j), by definition equals unity provided j is a common divisor of
n and 12, and equals zero otherwise. Keeping in mind φ(1) = 1 e.g., [23]
§24.3.2, Eq. (70) may be cast uder the form:
!
12
1 12
1 X
12/j
νM (n) =
+
ζ(12, j)ζ(n, j)φ(j)
n/j
12 n
12 j=2
!
;
(71)
where the first term on the right-hand side of Eq. (71) equals its counterpart
in Eq. (69) that is, via Eq. (7), the fractional number, NC /n, of distinct nchords. Then the second term on the right-hand side of Eq. (71) equals the
fractional number, ∆N/n, of repeating n-chords, as listed in Table 6.
Leaving aside the special case of null cardinality, n = 0, an inspection of
Table 6 shows the terms of the sum on the right-hand side of Eqs. (71) and
(69), respectively, coincide with the exception of n = 4, 6, 8, even if an equal
result is attained, which implies related formulations are not intrinsically
equivalent.
For n = 0, limn→0+ ∆N(n)/n = 11/12 is inferred from Table 1 via
Eqs. (14) and (69). On the other hand, (1+2+2+2+4)/12=11/12 via Eq. (71),
in agreement with the results listed in Table 6.
A similar comparison could be performed with regard to Tn /Tn I set
classes by use of the counterparts of Eqs. (69) and (70) e.g., [15] §72 which
implies a more complex formulation and for this reason it has not been considered.
30
Table 6: Fractional number of repeating n-chords, ∆N/n, of each cardinality,
n, 1 ≤ n ≤ 12, inferred from group theory (GT), Eq. (71), and from the
current paper (CP), Eq. (69). For n = 4, 6, 8, related addends are different
even if the sum yields equal results. For n = 0, Eq. (14) has also been used.
See text for further details.
n
0
1
2
3
4
5
6
7
8
9
10
11
12
φ ∆N/n (GT) ∆N/n (CP)
1
11/12
11/12
1
0/12
0/1
1
6/12
1/2
2
8/12
2/3
2
(15+6)/12
(3+4)/4
4
0/12
0/5
2 (20+12+4)/12 (5+4+9)/6
6
0/12
0/7
4
(15+6)/12
(6+8)/8
6
8/12
6/9
4
6/12
5/10
10
0/12
0/11
4
11/12
11/12
31
Symmetries found in enumerating n-chords, as shown by results listed in
Tables 1, 2, 3, are intrinsic to the geometrical interpretation. More specifically, n-hedrons exhibit (j − 1)-faces and (n − j − 1)-faces, 1 ≤ j ≤ n − 1,
in equal number, as listed in Table 7, which can be expressed via binomial
coefficients. In addition, the boundary condition expressed by Eq. (1) implies
regular, inclined n-hedrons, Ψn12,i ,where vertexes lie on the coordinate axes
at a distance, xi = 12, 1 ≤ i ≤ n, from the origin.
It is worth emphasizing symmetries are intrinsic to n-hedrons and, for this
reason, can be described using the binomial formula. In particular, regular
inclined n-hedrons, Ψn12 , can be partitioned into n congruent n-hedrons, Ψn12,i ,
1 ≤ i ≤ n, by joining the orthocentre with each vertex. Different Ψn12,i have
in common (n − 2)-faces where a vertex is the orthocentre of Ψn12 .
All of n-chords belonging to an assigned Tn or Tn /Tn I set class are similarly or similarly and symmetrically placed, respectively, within different
Ψn12,i , where distinct n-chords relate to positive integer points belonging to
Ψn12,i , 1 ≤ i ≤ n, while repeating n-chords, if any, are placed on (n − 2)-faces
of Ψn12,i where a vertex is the orthocentre of Ψn12 .
The geometric representation exploited in the current paper looks appealing, in that the idea is very simple and very effective, which is always a
good point. The fact that the dimension of the (n − 1)-plane where n-chords
are located only depends on the scale size, and not on the chromatic gamut
in which the scale is embedded (it only affects the number of points considered) makes it very useful. One could for example easily visualise triads in
quarter-tone scales, or on even finer-grained scales, on an ordinary (2-) plane.
Last but not least, the above mentioned geometric representation has musical significance: rotations around the n-sector (i.e. an axis normal to the
(n − 1)-plane through the orthocentre of Ψn12 ) correspond to chord inversions
(in musical language), or cyclic permutations (in mathematical language).
Looking at their location tells if a scale may have internal symmetries (repeating chords). Similarly, rotations about an axis joining orthocentres of
opposite (j − 1) and (n − 1 − j)-faces correspond to chord reflections, which
are symmetrically placed with respect to the rotation axis. But surely there
are many more which would be worth investigating further.
Enumeration of n-chords within an algebraic or geometrical framework
yields coinciding results but, in the latter alternative, further perspectives
can be exploited such as counting integer points within lattice polytopes.
Following a similar line of thought, the results of the current paper could
be extended to musical intervals in multiples of semitones under L-note equal
temperament, reproducing results found within the framework of group theory e.g., [17] [15].
The group theory, of course, remains the more powerful tool in dealing
32
with any kind of equivalence relation, but further insight on specific problems
could be gained in the light of both algebraic and geometric method outlined
in the current paper.
More specifically, using group theory in counting specified pitch-class sets
could be related to the description of a high-population (N ≫ 1) statistical
system e.g., a perfect gas within a box, in terms of observables e.g., volume,
pressure, temperature. Conversely, the enumeration of pitch-class sets within
an algebraic or geometric framework, as outlined in the current paper, could
>
be related to the description of a low-population (N ∼ 1) statistical system
e.g., perfectly elastic dissipationless spheres moving on a horizontal plane
bounded by perfectly elastic dissipationless walls, in terms of integration of
the motion equations from the knowledge of initial conditions.
5
Conclusion
The current investigation is restricted to musical intervals in multiples of
semitones under 12-note equal temperament, in particular about the question of how many n-chords there are of a given cardinality, and how many Tn
or Tn /Tn I set classes. In the light of a geometrical interpretation, n-chords
of a given cardinality, {ℓ1 , ℓ2 , ..., ℓn }, satisfying the boundary condition expressed by Eq. (1), are related to positive integer points in ℜn , belonging to a
regular inclined n-hedron, Ψn12 . Joining vertexes with the orthocentre yields n
congruent n-hedrons, Ψn12,i , 1 ≤ i ≤ n, where points with coordinates related
by circular permutations i.e. the n elements of Tn set classes, are equally
partitioned and similarly placed. If Tn set classes exhibit repeating n-chords,
then related points are lying on (n − 2)-faces of Ψn12,i where a vertex is the
orthocentre of Ψn12 .
The number, NC (n), of distinct n-chords is determined within an algebraic framework, and a symmetry is found around q = 13/2 as listed in Table
1. The fractional number, NC (n)/n, of distinct n-chords shows a symmetry
around n = 6, as listed in Table 1, provided the domain is extended to
n = 0 in connection with Euclidean 0-spaces, ℜ0 . The fractional numbers,
∆N(n)/n and NM (n)/n = [NC (n) + ∆N(n)]/n, of repeating and total (distinct + repeating) n-chords, respectively, also exhibit a symmetry around
n = 6, including n = 0, as listed in Table 1. It is worth remembering
νM (n) = NM (n)/n is the number of Tn set classes.
Following the same line of thought, the above results can be generalized to
musical intervals in multiples of semitones under L-note equal temperament.
In particular, Eqs. (5), (7), (8), (12), (13), (14), hold provided 12∓k, k = 0, 1,
is replaced by L ∓ k therein.
33
The above mentioned symmetries are disclosed within a geometrical framework, where n-chords together with Tn and Tn /Tn I set classes can be enumerated as well. More specifically, symmetries are intrinsic to n-hedrons, in
general on one hand, and via the boundary condition of the problem on the
other hand.
The main results of the current paper can be summarized as follows.
(1) Using a one-to-one correspondence between n-chords and positive integer
points within an Euclidean n-space, ℜn , both distinct and repeating n-chords
are enumerated within the framework of an algebraic method together with
Tn and Tn /Tn I set classes. Related results equal their counterparts already
known in the context of group theory e.g., [17] [14] Chap. 9 [15].
(2) Positive integer points related to n-chords of cardinality, n, belong to
a regular inclined n-hedron, Ψn12 , of vertexes, Vi ≡ (12δ1i , 12δ2i , ..., 12δni ),
1 ≤ i ≤ n. By joining the orthocentre, Hn ≡ (12/n, 12/n, ..., 12/n), with
vertexes, Vi , Ψn12 can be subdivided into n congruent n-hedrons, Ψn12,i , 1 ≤
i ≤ n. Positive integer points related to n-chords within an assigned Tn or
Tn /Tn I set class are equally partitioned among and similarly or similarly and
symmetrically placed within Ψn12,i . In particular, points related to repeating
n-chords are placed on (n − 2)-faces between different Ψn12,i , to be counted
according to their multiplicity i.e. the number of involved Ψn12,i .
(3) Both distinct and repeating n-chords are enumerated within the framework of a geometrical method together with Tn and Tn /Tn I set classes, reproducing their counterparts inferred from an algebraic method in (1).
(4) Results in (3) allow the calculation of nonnegative integer (i.e. exhibiting nonnegative integer coordinates) points belonging to Ψn12 which, to this
respect, can be considered as special cases of lattice polytopes in ℜn .
(5) Symmetries shown by the number of n-chords and Tn or Tn /Tn I set
classes with respect to the cardinality, n = 6, are intrinsic to the geometry
in a twofold manner. In general, the number of (j − 1)-faces of n-hedrons,
1 ≤ j ≤ n − 1, can be expressed via binomial coefficients as listed in Table 7.
In particular, the boundary condition of the problem, expressed by Eq. (1),
implies consideration of regular inclined n-hedrons, Ψn12 , where the number
of positive integer points can also be expressed via the binomial formula, as
shown in Table 1. To this respect, ℜn and ℜ12−n , 0 ≤ n ≤ 12, or ℜ6−m ,
ℜ6+m , 0 ≤ m ≤ 6, can be considered as Euclidean complementary spaces.
More generally, the symmetry of the results inferred from the group theory
could be conceived as intrinsic to lattice polytopes in ℜn .
34
Acknowledgements
The authors are indebted to two anonymous referees of an earlier version of
the paper (2008), for critical comments and useful suggestions.
References
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Press, Cambridge, 2006.
35
[15] J.
Hook,
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Twenty-Nine
Tetrachords?
A Tutorial on Combinatorics and Enumeration in Music
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der Musiktheorie, Bayreuther Mathematische Schriften, 45, 19-135, 1993.
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36
Appendix
A
Birthday-cake problem
Let N candles of L different colour be put at equal distance from the symmetry axis of a circular birthday cake, on the N vertexes of a regular polygon.
Accordingly, a generic configuration is defined by a N-tuple of colours, selected among the available L, in brief a colour set. Then Tn set classes
are defined by N repeated rotations of the cake about its axis by an angle,
αN = 2π/N.
The problem is: how many colour sets and Tn set classes do exist? For
sake of clarity, let attention be restricted to a simple case, N = 4 and L = 3:
blue (B), green (G), red (R), say e.g., [15] §46.
A solution to the problem shall be exploited along the following steps.
(i) Calculate the total number of distinct colour sets, 4-tuples in the case
under consideration.
(ii) Select different kinds of colour sets with regard to the elements of the
4-tuples.
(iii) For each kind of colour sets infer the number of repeating 4-tuples
within related Tn set classes.
(iv) Calculate the total number of distinct + repeating colour sets.
(v) Calculate the total number of Tn set classes.
Different configurations of the birthday cake can be related to different
colour sets. For each place of a 4-tuple, there are three possibilities for the
choice of a colour (B,G,R), yielding a total of 34 = 81 distinct 4-tuples. In
addition, the following types of colour sets can be defined: x4 , where all
colours are identical; x3 y, where three colours are identical and the remaining one is different; x2 y 2, where two different colours are identical in pairs;
x2 yz, where two colours are identical and the remaining two are different. In
general, x = B, G, R; x 6= y 6= z.
For x4 colour sets, {x, x, x, x}, the number of distinct
4-tuples is 4!/(4 · 3 ·
2) = 1, and the number of different elements, x, equals 31 = 3. Accordingly,
the total number of distinct and repeating 4-tuples is 3 · 1 = 3 and 3 · 3 = 9,
respectively. Then the total number of distinct + repeating colour sets is
3 + 9 = 12.
For x3 y colour sets, {x, x, x, y}, the number of distinct
4-tuples is 4!/(3 ·
2) = 4, and the number of different pairs, xy, equals 42 = 6. Accordingly,
the total number of distinct and repeating 4-tuples is 6 · 4 = 24 and 0 · 4 = 0,
37
respectively. Then the total number of distinct + repeating colour sets is
24 + 0 = 24.
For x2 y 2 colour sets, {x, x, y, y}, the number of distinct
4-tuples is 4!/(2 ·
2) = 6, and the number of different pairs, xy, equals 32 = 3. Accordingly,
the total number of distinct and repeating 4-tuples is 3·6 = 18 and 3·(1+1) =
6, respectively, the last due to {x, y, x, y} and {y, x, y, x}. Then the total
number of distinct + repeating colour sets is 18 + 6 = 24.
For x2 yz colour sets, {x, x, y, z}, the number of distinct
4-tuples is 4!/2 =
3
12, and the number of different triads, xyz, equals 1 = 3. Accordingly, the
total number of distinct and repeating 4-tuples is 3 · 12 = 36 and 0 · 12 = 0,
respectively. Then the total number of distinct + repeating colour sets is
36 + 0 = 36.
More specifically, in the case under discussion the total number of colour
sets belonging to an assigned type can be determined along the following
steps.
(a) Start from an assigned Tn set class of the type considered.
(b) Exchange y with z (if both present) to get a distinct (if any) Tn set class.
(c) Exchange x with y (only once in case of multiplicity) to get a distinct
(if any) Tn set class.
For x4 colour sets, the application of the method yields:
{x, x, x, x}, {x, x, x, x}, {x, x, x, x}, {x, x, x, x} ;
where no additional exchange has to be performed as both y and z are absent,
yielding 1 Tn set class made of 1 distinct and 3 repeating colour sets. Keeping
in mind x = B, G, R, there are 3 · 1 = 3 Tn set classes for a total of 1 · 3 = 3
distinct and 3 · 3 = 9 repeating colour sets, according to the above results.
For x3 y colour sets, the application of the method yields:
{x, x, x, y}, {x, x, y, x}, {x, y, x, x}, {y, x, x, x} ;
where no additional exchange has to be performed as z is absent, yielding 1 Tn set class made of 4 distinct colour sets. Keeping in mind xy =
BG, GR, RB, GB, RG, BR, (xy 6= yx due to different multiplicities in x and
y), there are 6 · 1 = 6 Tn set classes for a total of 6 · 4 = 24 distinct and
0 · 4 = 0 repeating colour sets, according to the above results.
For x2 y 2 colour sets, the application of the method yields:
{x, y, y, x}, {y, y, x, x}, {y, x, x, y}, {x, x, y, y} ;
{x, y, x, y}, {y, x, y, x}, {x, y, x, y}, {y, x, y, x} ;
38
where no additional exchange has to be performed as z is absent, yielding 2
Tn set classes made of (4+2) distinct and 2 repeating colour sets. Keeping
in mind xy = BG, GR, RB, (xy = yx do to equal multiplicities in x and y),
there are 3 · 2 = 6 Tn set classes for a total of 3 · (4 + 2) = 18 distinct and
3 · 2 = 6 repeating colour sets, according to the above results.
For x2 yz colour sets, the application of the method yields:
{x, x, y, z}, {x, y, z, x}, {y, z, x, x}, {z, x, x, y} ;
{x, x, z, y}, {x, z, y, x}, {z, y, x, x}, {y, x, x, z} ;
{x, y, x, z}, {y, x, z, x}, {x, z, x, y}, {z, x, y, x} ;
where no additional exchange has to be performed, yielding 3 Tn set classes
made of 3 · 4 = 12 distinct colour sets. Keeping in mind x = B, G, R;
x 6= y 6= z; there are 3 · 3 = 9 Tn set classes for a total of 3 · 12 = 36 distinct
and 0 · 12 = 0 repeating colour sets, according to the above results.
Finally, the number of distinct, NC , repeating, ∆N, total, NM , colour
sets, respectively, reads:
NC = NC (x4 ) + NC (x3 y) + NC (x2 y 2) + NC (x2 yz) = 3 + 24 + 18 + 36
= 81 ;
(72)
4
3
2 2
2
∆N = ∆N(x ) + ∆N(x y) + ∆N(x y ) + ∆N(x yz) = 9 + 0 + 6 + 0
= 15 ;
(73)
NM = NC + ∆N = 12 + 24 + 24 + 36 = 96 ;
(74)
and the number of Tn set classes, νM , is:
νM =
NM
12 24 24 36
=
+
+
+
= 3 + 6 + 6 + 9 = 24 ;
4
4
4
4
4
(75)
or 96/4=24.
Within the framework of group theory, the application of Burnside’s
lemma to birthday-cake problem yields e.g., [15] §46:
νM =
81 3 9 3
96
+ + + =
= 24 ;
4
4 4 4
4
(76)
which is in agreement with Eq. (75) but the addends are different.
Within the framework of group theory, the application of Polya’s theorem
to birthday-cake problem yields e.g., [15] §62:
νM = (1 + 1 + 1) + (1 + 1 + 1 + 1 + 1 + 1) + 2(1 + 1 + 1) + 3(1 + 1 + 1)
= 3 + 6 + 6 + 9 = 24 ;
(77)
39
which is in agreement with Eq. (75), exhibiting equal addends.
The result from Eq. (76) is based on Burnside’s lemma and does not involve weights on Tn set classes. On the other hand, the weighted method
given by Polya’s theorem is simply a refinement of the same idea, showing
exactly how the 24 Tn set classes are distributed among the various weights
e.g., [15] §63. For further details, an interested reader is addressed to specialized monographs e.g., [15] or textbooks e.g., [14] Chap. 9. The current
method exactly reproduces the results inferred via Polya’s theorem.
B
Necklace problem
Let N beads of L different colour be put at equal distance from the symmetry axis of a circular necklace, on the N vertexes of a regular polygon.
Accordingly, a generic configuration is defined by a N-tuple of colours, selected among the available L, in brief a colour set. Then Tn set classes are
defined by N repeated rotations of the necklace about its axis by an angle,
αN = 2π/N; Tn I set classes are similarly defined, but with the addition of
flipping over; Tn /Tn I set classes are the union of pairwise Tn and Tn I set
classes. The problem is: how many colour sets and Tn /Tn I set classes do
exist? For sake of clarity, let attention be restricted to a simple case, N = 4
and L = 3: blue (B), green (G), red (R), say e.g., [15] §47.
With regard to distinct colour sets and Tn set classes, the results coincide
with their counterparts of the birthday-cake problem, where the total number
equals 81 and 24, respectively. The total number of Tn I set classes, by
definition, also equals 24. What remains to be done is the enumeration of
repeating colour sets and Tn /Tn I set classes.
Among the eight colour sets within a generic Tn /Tn I set class, two quadruplets exhibit colour sets related via circular permutation and, in addition,
four doublets related via reflection. Accordingly, palindrome and pseudo
palindrome colour sets must be counted as repeating. More specifically, Tn
and Tn I i.e. pairwise Tn set classes made of distinct colour sets yield a single
Tn /Tn I set class, while single Tn set classes made of palindrome or pseudo
palindrome colour sets yield a single Tn /Tn I set class.
For x4 colour sets, {x, x, x, x}, the number of distinct
4-tuples is 4!/(4 · 3 ·
2) = 1, and the number of different elements, x, equals 31 = 3. Accordingly,
the total number of distinct and repeating 4-tuples is 3 · 1 = 3 and 3 · 3 = 9,
respectively. But colour sets are palindrome in the case under discussion,
which implies 4 additional 4-tuples for a total of 3 · (3 + 4) = 21 repeating
colour sets. Then the total number of distinct + repeating colour sets is
3+21 = 24.
40
For x3 y colour sets, {x, x, x, y}, the number of distinct
4-tuples is 4!/(3 ·
2) = 4, and the number of different pairs, xy, equals 42 = 6. Accordingly,
the total number of distinct and repeating 4-tuples is 6 · 4 = 24 and 0 · 4 =
0, respectively. But colour sets are pseudo palindrome in the case under
discussion, which implies 4 additional 4-tuples for a total of 6 · (0 + 4) = 24
repeating colour sets. Then the total number of distinct + repeating colour
sets is 24+24 = 48.
For x2 y 2 colour sets, {x, x, y, y}, the number of distinct
4-tuples is 4!/(2 ·
2) = 6, and the number of different pairs, xy, equals 32 = 3. Accordingly,
the total number of distinct and repeating 4-tuples is 3·6 = 18 and 3·(1+1) =
6, respectively, the last due to {x, y, x, y} and {y, x, y, x}. But colour sets are
palindrome or pseudo palindrome in the case under discussion, which implies
4 + 4 = 8 additional 4-tuples for a total of 3 · (2 + 8) = 30 repeating colour
sets. Then the total number of distinct + repeating colour sets is 18+30 =
48.
For x2 yz colour sets, {x, x, y, z}, the number of distinct
4-tuples is 4!/2 =
3
12, and the number of different triads, xyz, equals 1 = 3. Accordingly, the
total number of distinct and repeating 4-tuples is 3 · 12 = 36 and 0 · 12 = 0,
respectively. But colour sets, {x, y, x, z}, are pseudo palindrome in the case
under discussion, which implies 4 additional 4-tuples for a total of 3·(0+4) =
12 repeating colour sets. Then the total number of distinct + repeating colour
sets is 36 + 12 = 48.
More specifically, in the case under discussion the total number of colour
sets belonging to an assigned kind can be determined along the following
steps.
(a) Start from an assigned Tn /Tn I set class of the kind considered.
(b) Exchange y with z (if both present) to get a distinct (if any) Tn /Tn I set
class.
(c) Exchange x with y (only once in case of multiplicity) to get a distinct
(if any) Tn /Tn I set class.
For x4 colour sets, the application of the method yields:
{x, x, x, x}, {x, x, x, x}, {x, x, x, x}, {x, x, x, x} ,
{x, x, x, x}, {x, x, x, x}, {x, x, x, x}, {x, x, x, x} ;
where no additional exchange has to be performed as both y and z are absent,
yielding 1 Tn /Tn I set class made of 1 distinct and 3+4 = 7 repeating colour
sets. Keeping in mind x = B, G, R, there are 3 · 1 = 3 Tn /Tn I set classes for
41
a total of 1 · 3 = 3 distinct and 7 · 3 = 21 repeating colour sets, according to
the above results.
For x3 y colour sets, the application of the method yields:
{x, x, x, y}, {x, x, y, x}, {x, y, x, x}, {y, x, x, x} ,
{y, x, x, x}, {x, y, x, x}, {x, x, y, x}, {x, x, x, y} ;
where no additional exchange has to be performed as z is absent, yielding 1
Tn /Tn I set class made of 4 distinct and 4 repeating colour sets. Keeping in
mind xy = BG, GR, RB, GB, RG, BR, (xy 6= yx due to different occurrences
in x and y), there are 6 · 1 = 6 Tn /Tn I set classes for a total of 6 · 4 = 24
distinct and 6 · 4 = 24 repeating colour sets, according to the above results.
For x2 y 2 colour sets, the application of the method yields:
{x, y, y, x}, {y, y, x, x}, {y, x, x, y}, {x, x, y, y}
{x, y, y, x}, {x, x, y, y}, {y, x, x, y}, {y, y, x, x}
{x, y, x, y}, {y, x, y, x}, {x, y, x, y}, {y, x, y, x}
{y, x, y, x}, {x, y, x, y}, {y, x, y, x}, {x, y, x, y}
,
;
,
;
where no additional exchange has to be performed as z is absent, yielding 2
Tn /Tn I set classes made of 4+2 = 6 distinct and 4+6 = 10 repeating colour
sets. Keeping in mind xy = BG, GR, RB, (xy = yx due to equal occurrences
in x and y), there are 3 · 2 = 6 Tn /Tn I set classes for a total of 3 · (4 + 2) = 18
distinct and 3 · (4 + 6) = 30 repeating colour sets, according to the above
results.
For x2 yz colour sets, the application of the method yields:
{x, x, y, z}, {x, y, z, x}, {y, z, x, x}, {z, x, x, y}
{z, y, x, x}, {x, z, y, x}, {x, x, z, y}, {y, x, x, z}
{x, x, z, y}, {x, z, y, x}, {z, y, x, x}, {y, x, x, z}
{y, z, x, x}, {x, y, z, x}, {x, x, y, z}, {z, x, x, y}
{x, y, x, z}, {y, x, z, x}, {x, z, x, y}, {z, x, y, x}
{z, x, y, x}, {x, z, x, y}, {y, x, z, x}, {x, y, x, z}
,
;
,
;
,
;
where no additional exchange has to be performed, yielding 2 Tn /Tn I set
classes (the first two coincide) made of 3 · 4 = 12 distinct and 1 · 4 = 4
repeating colour sets. Keeping in mind x = B, G, R, x 6= y 6= z, there are
3 · 2 = 6 Tn /Tn I set classes for a total of 3 · 12 = 36 distinct and 3 · 4 = 12
repeating colour sets, according to the above results.
42
Finally, the total number of distinct, NC , repeating, ∆N, and total, NM ,
colour sets, respectively, reads:
NC = NC (x4 ) + NC (x3 y) + NC (x2 y 2 ) + NC (x2 yz) = 3 + 24 + 18 + 36
= 81 ;
(78)
4
3
2 2
2
∆N = ∆N(x ) + ∆N(x y) + ∆N(x y ) + ∆N(x yz) = 21 + 24 + 30 + 12
= 87 ;
(79)
NM = NC + ∆N = 24 + 48 + 48 + 48 = 168 ;
(80)
and the number of Tn /Tn I set classes, νM , is:
νM =
NM
24 48 48 48
=
+
+
+
= 3 + 6 + 6 + 6 = 21 ;
8
8
8
8
8
(81)
or 168/8=21.
Within the framework of group theory, the application of Burnside’s
lemma to necklace problem yields e.g., [15] §47:
νM =
168
84 12 36 36
+
+
+
=
= 21 ;
8
8
8
8
8
(82)
which is in agreement with Eq. (81) but the addends are different.
Within the framework of group theory, the application of Polya’s theorem
to necklace problem yields e.g., [15] §64:
νM = (1 + 1 + 1) + (1 + 1 + 1 + 1 + 1 + 1) + 2(1 + 1 + 1) + 2(1 + 1 + 1)
= 3 + 6 + 6 + 6 = 21 ;
(83)
which is in agreement with Eq. (81), exhibiting equal addends.
The result from Eq. (82) is based on Burnside’s lemma and does not involve weights on Tn /Tn I set classes. On the other hand, the weighted method
given by Polya’s theorem is simply a refinement of the same idea, showing exactly how the 21 Tn /Tn I set classes are distributed among the various weights
e.g., [15] §63. For further details, an interested reader is addressed to specialized monographs e.g., [15] or textbooks e.g., [14] Chap. 9. The current
method exactly reproduces the results inferred via Polya’s theorem.
C
C.1
Basic ideas on Ψn12 n-hedrons
Analytic geometry in ℜk
The formulation of analytic geometry in ordinary Euclidean space, ℜ3 , can
easily be extended to Euclidean k-spaces, ℜk .
43
With regard to a Cartesian orthogonal reference frame, (O x1 x2 , ... xk ),
let PA ≡ (xA1 , xA2 , ..., xAk ) and PB ≡ (xB1 , xB2 , ..., xBk ) be generic points.
The related square distance reads:
2
(PA PB ) =
k
X
i=1
(xAi − xBi )2 ;
(84)
if PA is fixed (the origin say) while PB remains generic, then Eq. (84) represents a k-dimension sphere, or k-sphere, of radius, R = PA PB , centered on
PA .
The mean point, PM , of the segment joining the points, PA , PB , reads:
xA1 + xB1 xA2 + xB2
xAk + xBk
PM ≡
,
, ...,
2
2
2
;
(85)
and the identities, PA PM = PB PM , PA PM + PB PM = PA PB , can easily be
verified via Eqs. (84) and (85).
Let r, r ′ , be generic straight lines in ℜk , expressed as:
x1 − x10
x2 − x20
xk − xk0
=
= ... =
;
L1
L2
Lk
1≤i≤k ;
Li = xi0 − xi0 ;
′
′
′
′
x1 − x10
x2 − x20
x′k − x′k0
;
=
=
...
=
L′1
L′2
L′k
1≤i≤k ;
L′i = x′i0 − x′i0 ;
r:
r′ :
(86)
(87)
(88)
(89)
where P0 ≡ (x10 , x20 , ..., xk0 ), P0 ≡ (x10 , x20 , ..., xk0 ), are selected points on r
′
and P′0 ≡ (x′10 , x′20 , ..., x′k0 ), P0 ≡ (x′10 , x′20 , ..., x′k0 ), are selected points on r ′ .
c′ , can be expressed as:
The angle, rr
c′ | =
| cos rr
k
X
′
Li Li
i=1
" k
#1/2 #1/2 " k
X
X
(L′i )2
L2i
;
(90)
i=1
i=1
c′ = 0.
where the orthogonality condition reads cos rr
k
Let p be a generic (k − 1)-plane in ℜ , expressed as:
p:
a1 x1 + a2 x2 + ... + ak xk + a0 = 0 ;
(91)
where −a0 /ai , is the intercept of p on the coordinate axis, xi , 1 ≤ i ≤ k.
44
The angle, rcp, between the straight line, r, and the (k − 1)-plane, p, can
be expressed as:
| sin rcp| =
k
X
Li ai
i=1
" k
#1/2 #1/2 " k
X
X
a2i
L2i
;
(92)
i=1
i=1
where the orthogonality condition reads sin rcp = ∓1.
For further details, an interested reader is addressed to earlier investigations [24] Chap. 2 §2.16 [25] [26] [27] Chap. 4 §4.17.
C.2
General properties of (k + 1)-hedrons
With regard to an Euclidean k-space, ℜk , and a Cartesian orthogonal reference frame, (Ox1 x2 ... xk ), let (k + 1) k-misaligned points be assigned as
V1 , V2 , ..., Vk+1 . Points are k-misaligned in the sense different k-tuples,
{Vi1 , Vi2 , ..., Vik }, 1 ≤ i1 < i2 < ... < ik ≤ k + 1, belong to different (k − 1)planes in ℜk . Let each point be connected with the remaining ones by segments. Let the resulting geometrical figure be defined as (k + 1)-hedron, and
denoted as Φk+1 . In ordinary space, k = 3, 4-hedrons reduce to ordinary
tetrahedrons.
The above definition could appear ambiguous, in that the etymology of
tetrahedron implies the existence of four faces. Keeping in mind (k + 1)hedrons have k-dimensions, it can be seen (as shown below) that the number
of (k − 1)-faces equals the number of connected points, (k + 1). Then etymological meaning is preserved provided (k − 1)-faces are considered instead
of ordinary (2-) faces.
Another source of ambiguity lies in the fact, that “(k + 1)-hedrons” are
already defined in ordinary space, ℜ3 , concerning ordinary faces e.g., hexahedron, octahedron, dodecahedron, icosahedron. Throughout the present
paper, according to the current definition, (k + 1)-hedrons are intended as
solids in ℜk where a generic point, Vi , 1 ≤ i ≤ k + 1, is connected to the
remaining (k-misaligned) k.
Any point, Vi , 1 ≤ i ≤ k + 1, is a vertex or 0-face of Φk+1 . Accordingly,
the
number of 0-faces equals the number of distinct 1-tuples of vertexes,
k+1
= (k + 1)/1!.
1
Any duo of vertexes, Vi Vj , i 6= j, yields a side or 1-face of Φk+1 . Accordingly,thenumber of 1-faces equals the number of distinct 2-tuples of
vertexes, k+1
= (k + 1)k/2!.
2
45
Any trio of vertexes, Vi Vj Vℓ , i 6= j 6= ℓ, yields a triangle or 2-face of Φk+1 .
Accordingly,
the
number of 2-faces equals the number of distinct 3-tuples of
k+1
vertexes, 3 = (k + 1)k(k − 1)/3!.
Any quartet of vertexes, Vi Vj Vℓ Vm , i 6= j 6= ℓ 6= m, yields a tetrahedron
or 3-face of Φk+1 . Accordingly,
the
number of 3-faces equals the number of
k+1
distinct 4-tuples of vertexes, 4 = (k + 1)k(k − 1)(k − 2)/4!.
In general, any j-tet of vertexes, Vi1 Vi2 ...Vij , i1 6= i2 6= ... 6= ij , 1 ≤ j ≤ k,
yields a j-hedron or (j − 1)-face of Φk+1 . Accordingly, the
of (j − 1) number
k+1
faces equals the number of distinct j-tuples of vertexes, j = (k+1)k...(k−
j + 2)/j!.
To this respect, a useful recursion formula follows from the identity:
!
!
k
k
k+1
+
=
j−1
j
j
!
;
1≤j≤k ;
(93)
which translates into:
Fj (Φk+1 ) = Fj (Φk ) + Fj−1 (Φk ) ;
1≤j≤k ;
(94)
where, in general, Fℓ (Φm+1 ) is the number of (ℓ−1)-faces of Φm+1 , 1 ≤ ℓ ≤ m.
C.3
A special case: Ψn12 regular inclined n-hedrons
With regard to an Euclidean n-space, ℜn , and a Cartesian orthogonal reference frame, (Ox1 x2 ... xn ), let a regular, inclined, n-hedron where vertexes
are placed on the coordinate axes at a distance, xi = 12, 1 ≤ i ≤ 12, from
the origin, be considered and denoted as Ψn12 . Points with positive integer
coordinates, or positive integer points, within Ψn12 satisfy Eq. (1) as shown in
the text. More specifically, the extension of Eq. (1) to real coordinates reads:
pn :
x1 + x2 + ... + xn = 12 ;
1 ≤ n ≤ 12 ;
(95)
which represents a (n − 1)-plane where the intercepts on the coordinate axes
coincide with the vertexes of Ψn12 .
The n-sector (n = 2, bisector; n = 3, trisector; and so on) of the positive
n
2 -ant, by definition, includes both the origin and the point of unit coordinates, which implies xi0 = 0, xi0 = 1, 1 ≤ i ≤ n, and Eq. (86) reduces
to:
rn :
x1 = x2 = ... = xn ;
1 ≤ n ≤ 12 ;
which is the locus of points with identical coordinates.
46
(96)
The angle between the (n − 1)-plane, pn , and the n-sector, rn , keeping in
mind ai = 1, Li = 1, 1 ≤ i ≤ n, via Eq. (92) reads:
| sin rd
n pn | =
n √ √ n n
=1 ;
(97)
which implies rn is orthogonal to pn through the orthocentre of Ψn12 that is,
in fact, the sole point of pn with identical coordinates.
With regard to the generic (j − 1)-face, 1 ≤ j ≤ n, belonging to Ψn12 , the
orthocentre, Hj , exhibits j equal coordinates in that it is equally distant from
j vertexes of Ψn12 . Accordingly, nonzero coordinates equal 12/j via Eq. (1).
The result is:

Hj (i1 , i2 , ..., ij ) ≡ 
j
X
j
X
j
X

12
12
12
δik 1 ,
δik 2 , ...,
δi j  ;
j k=1
j k=1
j k=1 k
1 ≤ i1 < i2 < ... < ij ≤ n ;
(98)
where regular vertexes, regular sides, regular triangles, regular tetrahedrons,
and so on, are conceived as 0-faces, 1-faces, 2-faces, 3-faces, and so on, respectively. In particular, the orthocentre of Ψn12 , conceived as related to a
single (n − 1)-face, reads:
12
12 12
;
(99)
, , ...,
Hn (1, 2, ..., n) ≡
n n
n
where Hn is a positive integer point only if the ratio, 12/n, is integer, which
implies n = 1, 2, 3, 4, 6, 12.
The number of (j − 1)-faces, Fj , 1 ≤ j ≤ n, belonging to Ψn12 , 1 ≤ n ≤ 13,
can be inferred from the general results mentioned above, where k = n−1, as
listed in Table 7. More specifically, defining (−1)-faces and (n − 1)-faces as
n-hedron metacentres and n-hedrons, respectively, related values equal unity
and Table 7 reproduces Pascal’s triangle e.g., [15] §30.
The metacentre of Ψn12 is defined as the centre of a n-sphere, on the surface
of which vertexes of Ψn12 are lying i.e. the origin of coordinates. In addition,
the metacentre can be thought of as the orthocentre of a (n + 1)-hedron
where Ψn12 is a (n − 1)-face and the additional vertex lies on the n-sector of
the negative 2n -ant, at a distance, Rn = 12, from the origin.
In the special case, n = 0, related 0-hedron lies outside ℜ0 and no vertex
can be defined, while the metacentre coincides with ℜ0 , yielding 1 (−1)-face.
Let a (j−1)-face and a (n−j−1)-face of Ψn12 be assigned. The coordinates
of generic points lying within, Pj , Pn−j , respectively, read:

Pj ≡ ℓ1
j
X
k=1
δik 1 , ℓ2
j
X
δik 2 , ..., ℓn
k=1
j
X
k=1
47

δik j  ;
Table 7: Number of (j−1)-faces, Fj , 0 ≤ j ≤ n, belonging to Ψn12 , 0 ≤ n ≤ 13.
In particular, n = F1 . The addition of (−1)-faces, F0 , related to n-hedron
metacentres, and (n−1)-faces, Fn , related to n-hedrons, makes the horizontal
and oblique unit line, respectively, yielding Pascal’s triangle. For j < 10, Fj
is denoted as F0j to save aesthetics. See text for further details.
F00
F01
F02
F03
F04
F05
F06
F07
F08
F09
F10
F11
F12
F13
1
1 1 1
1 2 3
1 3
1
1 1 1
4 5 6
6 10 15
4 10 20
1 5 15
1 6
1
1
7
21
35
35
21
7
1
48
1
1
1
1
1
1
8
9 10 11 12
13
28 36 45 55 66
78
56 84 120 165 220 286
70 126 210 330 495 715
56 126 252 462 792 1287
28 84 210 462 924 1716
8 36 120 330 792 1716
1
9 45 165 495 1287
1 10 55 220 715
1 11 66 286
1 12
78
1
13
1
1 ≤ i1 < i2 < ... < ij ≤ n ;

Pn−j ≡ ℓ′1
n−j
X
δi′m 1 , ℓ′2
1≤
δi′m 2 , ..., ℓ′n
<
i′2
< ... <
n−j
X
m=1
m=1
m=1
i′1
n−j
X
i′n−j
≤n ;
(100a)

δi′m j  ;
(100b)
and the hyperfaces under discussion are defined as opposite if, in addition,
ik 6= i′m , 1 ≤ k ≤ j, 1 ≤ m ≤ n − j.
The special case, Pj ≡ Hj , Pn−j ≡ Hn−j , via Eq. (98) reads:


j
j
j
12 X
12 X
12 X
Hj ≡ 
δik 1 ,
δik 2 , ...,
δi n  ;
j k=1
j k=1
j k=1 k
1 ≤ i1 < i2 < ... < ij ≤ n ;

Hn−j ≡ 
(101a)

X
X
X
12 n−j
12 n−j
12 n−j
δi′m 1 ,
δi′m 2 , ...,
δi′ n  ;
n − j m=1
n − j m=1
n − j m=1 m
1 ≤ i′1 < i′2 < ... < i′n−j ≤ n ;
(101b)
n
where ik 6= i′m ; 1 ≤ k ≤ j; 1 ≤ m ≤ n − j; and the orthocentre of ψ12
, Hn , is
necessarily aligned with Hj and Hn−j .
More specifically, the particularization of Eq. (86) to xi0 = (12/j)(δi1 i +
δi2 i + ... + δij i ), xi0 = [12/(n − j)](δi′1 i + δi′2 i + ... + δi′n−j i ), 1 ≤ i ≤ n, yields:
x1 −
j
12 X
δi 1
j k=1 k
j
X
12 n−j
12 X
δi′m 1 −
δi 1
n − j m=1
j k=1 k
=
=
x2 −
j
12 X
δi 1
j k=1 k
j
X
12 n−j
12 X
δi′m 1 −
δi 1
n − j m=1
j k=1 k
xn −
j
12 X
δi n
j k=1 k
j
X
12 n−j
12 X
δi′m n −
δi n
n − j m=1
j k=1 k
= ...
;
(102)
where, in the case under discussion of opposite hyperfaces, each sum equals
zero or unity and attains different values on different points. Accordingly,
the generic term appearing in Eq. (102) reads either −jxi /12 + 1, for a total
of j occurrences, or (n − j)xi /12, for a total of (n − j) occurrences, 1 ≤ i ≤ n.
In the special case of the orthocentre of Ψn12 , xi = 12/n, 1 ≤ i ≤ n, and the
above mentioned alternatives reduce to either −(j/12)(12/n) + 1 = (n − j)/n
or [(n − j)/12](12/n) = (n − j)n, which implies Eq. (102) is satisfied and,
49
in turn, the orthocentre of Ψn12 is aligned with the orthocentre of opposite
(j − 1) and (n − j − 1)-faces.
Let Pn ≡ (ℓ1 , ℓ2 , ..., ℓn ) be a generic integer point belonging to Ψn12 . The
square distance, (Pn Hn )2 = Rn2 , from the orthocentre of Ψn12 , via Eqs. (84)
and (99) reads:
n
X
12
Rn2 = (ℓi − ℓ0 )2 ;
ℓ0 =
;
(103)
n
i=1
where Rn is the radius of a (n − 1)-circle centered on Hn .
The straight line joining Pn and Hn , via Eq. (86) is expressed as:
ra :
x1 − ℓ0
x2 − ℓ0
xn − ℓ0
=
= ... =
;
ℓ1 − ℓ0
ℓ2 − ℓ0
ℓn − ℓ0
(104)
x2 − ℓ0
xn − ℓ0
x1 − ℓ0
;
=
= ... =
ℓ i1 − ℓ 0
ℓ i2 − ℓ 0
ℓ in − ℓ 0
(105)
where Lk = ℓk − ℓ0 , 1 ≤ k ≤ n.
The counterpart of Eq. (104), related to a selected permutation of the
coordinates of Pn , is:
rb :
where Lik = ℓik − ℓ0 , 1 ≤ k ≤ n.
The angle, rd
a rb , via Eq. (90) reads:
| cos rd
a rb | =
n
X
(ℓk − ℓ0 )(ℓik − ℓ0 )
k=1
" k
#1/2 #1/2 " k
X
X
(ℓik − ℓ0 )2
(ℓk − ℓ0 )2
;
(106)
i=1
i=1
where, in the case under discussion, the sums of squares exhibit same addends
in different order. Accordingly, Eq. (106) reduces to:
| cos rd
a rb | =
n
X
(ℓ
−
ℓ
)(ℓ
−
ℓ
)
k
0
i
0
k
k=1
k
X
(ℓk − ℓ0 )2
;
(107)
i=1
where the permutation of the coordinates acts only on the sum of products.
The special case of circular permutation, ik = k + 1, reads:
| cos rd
a rb | =
n
X
(ℓk − ℓ0 )(ℓk+1 − ℓ0 ) k=1
k
X
2
(ℓk − ℓ0 )
i=1
50
;
(108)
where ℓn+1 = ℓ1 . It is apparent adjacent circular permutations of coordinates
e.g., (ℓ1 , ℓ2 , ..., ℓn ), (ℓ2 , ℓ3 , ..., ℓ1 ); (ℓ2 , ℓ3 , ..., ℓ1 ), (ℓ3 , ℓ4 , ..., ℓ2 ); yield the same
angle in that the sum of products in Eq. (108) exhibits same addends in
different order. On the other hand, non adjacent circular permutations of
coordinates yield different angles with respect to the sum of adjacent angles,
as related straight lines are not complanar. In fact, positive integer points
whose coordinates belong to a same Tn set class are similarly placed within
different Ψn12,i , 1 ≤ i ≤ n, i.e. congruent (n−1)-hedrons, where the ith vertex
coincides with the orthocentre of Ψn12 and the remaining ones coincide with
vertexes (different from the ith) of Ψn12 .
The special case of reflection, ik = n − k + 1, reads:
| cos rd
a rb | =
n
X
(ℓk − ℓ0 )(ℓn−k+1
k=1
k
X
(ℓk − ℓ0 )2
i=1
− ℓ0 ) ;
(109)
where adjacent circular permutations yield the same angle as the sum of
products in Eq. (109) exhibits same addends in different order. In fact, positive integer points with coordinates belonging to a same Tn /Tn I set class are
similarly and symmetrically placed within different Ψn12,i , 1 ≤ i ≤ n.
The mean point, Mn , of the segment joining pairwise points, Pn ≡ (ℓ1 , ℓ2 , ..., ℓn ),
Qn ≡ (ℓn , ℓn−1, ..., ℓ1 ), via Eq. (85) reads:
Mn ≡
ℓ1 + ℓn ℓ2 + ℓn−1
ℓn + ℓ1
,
, ...,
,
2
2
2
!
;
(110)
accordingly, the coordinates of Mn are palindrome as either (a1 , ..., an/2 , an/2 , ..., a1 )
or (a1 , ..., a(n−1)/2 , a(n+1)/2 , a(n−1)/2 , ..., a1 ) according if n is even or odd, respectively.
To get further insight, let Ψn12 be conceived as a series of Chinese-box
n
n
n-hedronical shells, ψk,12
, where the external shell, ψ0,12
, has not to be considered in that related points exhibit one null coordinate at least. The last
shell of the series relates to the value of k for which the condition, k ≤ 12/n,
n
still holds. The special case, k = 12/n, implies ψ12/n,12
coincides with the
n
orthocentre of Ψ12 , provided n is a divisor of 12. For instance, vertexes of
n
ψk,12
read Vi′ ≡ [k + δi1 (12 − kn), k + δi2 (12 − kn), ..., k + δin (12 − kn)], and
mean points between vertexes read M′ij ≡ [k + (δi1 + δj1 )(12 − kn)/2, k + (δi2 +
δj2 )(12 − kn)/2, ..., k + (δin + δjn )(12 − kn)/2].
n
For even n, the straight line joining mean points on opposite side of ψk,12
,
51
restricting to palindrome coordinates, via Eq. (86) is expressed as:
xn/2−1 − k
xn/2 − [k + (12 − kn)/2]
=
[k + (12 − kn)/2] − k
k − [k + (12 − kn)/2]
xn/2+1 − [k + (12 − kn)/2]
xn/2+2 − k
=
=
;
k − [k + (12 − kn)/2]
[k + (12 − kn)/2] − k
(111)
which, after some algebra, reduces to:
xn/2−1 = −xn/2 + 2k + 6 −
kn
kn
= −xn/2+1 + 2k + 6 −
= xn/2+2 ; (112)
2
2
while the remaining coordinates are fixed as:
x1 = ... = xn/2−2 = xn/2+3 = ... = xn = k ;
(113)
n
in connection with the selected ψk,12
.
In conclusion, positive integer points exhibiting palindrome coordinates,
for even n lie on the straight line, expressed by Eqs. (112)-(113). For instance,
let n = 6, k = 1, be considered. Accordingly, Eqs. (112)-(113) reduce to:
x2 = −x3 + 5 = −x4 + 5 = x5 ;
x1 = x6 = 1 ;
(114)
where (1, 4, 1, 1, 4, 1), (1, 1, 4, 4, 1, 1), are coordinates of mean points on opposite sides of related 4-face, and (1, 3, 2, 2, 3, 1), (1, 2, 3, 3, 2, 1), are palindrome
coordinates of positive integer points on the straight line joining the above
mentioned mean points, expressed by Eq. (114).
For odd n, the straight line joining a vertex with the mean point of the
n
opposite sides of a 3-face of ψk,12
, restricting to palindrome coordinates, via
Eq. (86) is expressed as:
x(n−1)/2 − k
x(n+1)/2 − [k + (12 − kn)]
=
[k + (12 − kn)/2] − k
k − [k + (12 − kn)/2]
x(n+3)/2 − k
;
=
[k + (12 − kn)/2] − k
(115)
which, after some algebra, reduces to:
2x(n−1)/2 = −x(n+1)/2 + 12 − kn + 3k = 2x(n+3)/2 ;
(116)
while the remaining coordinates are fixed as:
x1 = ... = x(n−3)/2 = x(n+5)/2 = ... = xn = k ;
52
(117)
n
in connection with the selected ψk,12
.
In conclusion, positive integer points exhibiting palindrome coordinates,
for odd n lie on the straight line, expressed by Eqs. (116)-(117). For instance,
let n = 5, k = 1, be considered. Accordingly, Eqs. (116)-(117) reduce to:
2x2 = −x3 + 10 = 2x4 ;
x1 = x5 = 1 ;
(118)
where (1, 1, 8, 1, 1), (1, 9/2, 1, 9/2, 1), are coordinates of the vertex and mean
point on opposite side of related 3-face, respectively, and (1, 2, 6, 2, 1), (1, 3, 4, 3, 1),
(1, 4, 2, 4, 1), are palindrome coordinates of positive integer points on the
straight line joining the above mentioned vertex and mean point, expressed
by Eq. (118).
53
Figure 1: The 1 : 1 correspondence between projected points, Pn−1 ↔ Qn−1 ,
in the special case of an Euclidean n-space, n = 3, P2 (ℓ1 , ℓ2 ) ↔ Q2 (12−ℓ1 , ℓ2 ),
yielding a total number of distinct n-chords, NC (3) = 55. The coordinates
of projected points, P2 , satisfying Eq. (2) i.e. ℓ3 = 12 − ℓ1 − ℓ2 , are placed
above the straight line with positive slope. The coordinates of projected
points, Q2 ↔ P2 , are placed below the straight line with negative slope. The
coordinates of projected points, P2 ↔ Q2 , P′2 ↔ Q′2 , P2 ≡ Q′2 , P′2 ≡ Q2 ,
are placed within the left angle bisected by a horizontal straight line. The
coordinates of points for which the correspondence does not hold and Eq. (2)
is violated i.e. ℓ3 > 12 − ℓ1 − ℓ2 , are placed within the right angle bisected by
a horizontal line. In representing coordinates, brackets have been omitted to
save space. The correspondence, P2 ↔ Q2 , acts along columns. Initial zeroes
have been preferred to blank spaces to save aesthetics.
54
Figure 2: The special case, n = 3, where the regular inclined n-hedron
reduces to a regular triangle and congruent n-hedrons to isosceles triangles
with adjacent equal sides. The common vertex of isosceles triangles coincides
with the orthocentre of the regular triangle.
55
Figure 3: Representation of distinct n-chords of cardinality, n, as positive
integer coordinates of points in an Euclidean n-space, lying within a regular
inclined n-hedron via Eq. (1), in the special case, n = 3. The total number
of distinct 3-chords is NC (3) = 55. In representing coordinates, brackets
have been omitted to save space and ten has been replaced by zero to save
aesthetics.
56